Solved on Jan 17, 2024

Solve the quadratic equation $2n^2 = 27 - 15n$ for the value of $n$ .

STEP 1

Assumptions
1. The equation to solve is $2n^2 = 27 - 15n$ .
2. We will solve for $n$ by rearranging the equation into standard quadratic form and then using the quadratic formula or factoring, if possible.

STEP 2

First, we need to rearrange the equation into the standard quadratic form, which is $ax^2 + bx + c = 0$ .
$2n^2 = 27 - 15n$

STEP 3

Move all terms to one side of the equation to set it equal to zero.
$2n^2 + 15n - 27 = 0$

STEP 4

Now that we have the quadratic equation in standard form, we can attempt to factor it. If factoring is not possible or not easily seen, we can use the quadratic formula.

STEP 5

Check if the quadratic can be factored by finding two numbers that multiply to $2 \times (-27) = -54$ and add up to $15$ .

STEP 6

After checking for possible factors, we see that there are no two integers that satisfy both conditions. Therefore, we will use the quadratic formula to solve for $n$ .

STEP 7

Recall the quadratic formula:
$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

STEP 8

Identify the coefficients $a$ , $b$ , and $c$ from the quadratic equation $2n^2 + 15n - 27 = 0$ .
$a = 2, \quad b = 15, \quad c = -27$

STEP 9

Substitute the coefficients into the quadratic formula.
$n = \frac{-15 \pm \sqrt{15^2 - 4 \cdot 2 \cdot (-27)}}{2 \cdot 2}$

STEP 10

Calculate the discriminant ( $b^2 - 4ac$ ).
$15^2 - 4 \cdot 2 \cdot (-27) = 225 + 216 = 441$

STEP 11

Take the square root of the discriminant.
$\sqrt{441} = 21$

STEP 12

Substitute the square root of the discriminant back into the quadratic formula.
$n = \frac{-15 \pm 21}{4}$

STEP 13

Solve for the two possible values of $n$ .
$n_1 = \frac{-15 + 21}{4} = \frac{6}{4} = \frac{3}{2}$
$n_2 = \frac{-15 - 21}{4} = \frac{-36}{4} = -9$

STEP 14

The two solutions for $n$ are $n_1 = \frac{3}{2}$ and $n_2 = -9$ .
The solutions to the equation $2n^2 = 27 - 15n$ are $n = \frac{3}{2}$ and $n = -9$ .

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Solve the quadratic equation $2n^2 = 27 - 15n$ for the value of $n$ .

STEP 1

Assumptions
1. The equation to solve is $2n^2 = 27 - 15n$ .
2. We will solve for $n$ by rearranging the equation into standard quadratic form and then using the quadratic formula or factoring, if possible.

STEP 2

First, we need to rearrange the equation into the standard quadratic form, which is $ax^2 + bx + c = 0$ .
$2n^2 = 27 - 15n$

STEP 3

Move all terms to one side of the equation to set it equal to zero.
$2n^2 + 15n - 27 = 0$

STEP 4

Now that we have the quadratic equation in standard form, we can attempt to factor it. If factoring is not possible or not easily seen, we can use the quadratic formula.

STEP 5

Check if the quadratic can be factored by finding two numbers that multiply to $2 \times (-27) = -54$ and add up to $15$ .

STEP 6

After checking for possible factors, we see that there are no two integers that satisfy both conditions. Therefore, we will use the quadratic formula to solve for $n$ .

STEP 7

Recall the quadratic formula:
$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

STEP 8

Identify the coefficients $a$ , $b$ , and $c$ from the quadratic equation $2n^2 + 15n - 27 = 0$ .
$a = 2, \quad b = 15, \quad c = -27$

STEP 9

Substitute the coefficients into the quadratic formula.
$n = \frac{-15 \pm \sqrt{15^2 - 4 \cdot 2 \cdot (-27)}}{2 \cdot 2}$

STEP 10

Calculate the discriminant ( $b^2 - 4ac$ ).
$15^2 - 4 \cdot 2 \cdot (-27) = 225 + 216 = 441$

STEP 11

Take the square root of the discriminant.
$\sqrt{441} = 21$

STEP 12

Substitute the square root of the discriminant back into the quadratic formula.
$n = \frac{-15 \pm 21}{4}$

STEP 13

Solve for the two possible values of $n$ .
$n_1 = \frac{-15 + 21}{4} = \frac{6}{4} = \frac{3}{2}$
$n_2 = \frac{-15 - 21}{4} = \frac{-36}{4} = -9$

STEP 14

The two solutions for $n$ are $n_1 = \frac{3}{2}$ and $n_2 = -9$ .
The solutions to the equation $2n^2 = 27 - 15n$ are $n = \frac{3}{2}$ and $n = -9$ .

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Solve the quadratic equation 2n2=27−15n2n^2 = 27 - 15n2n2=27−15n for the value of nnn.

STEP 1

Assumptions1. The equation to solve is 2n2=27−15n2n^2 = 27 - 15n2n2=27−15n.2. We will solve for nnn by rearranging the equation into standard quadratic form and then using the quadratic formula or factoring, if possible.

STEP 2

First, we need to rearrange the equation into the standard quadratic form, which is ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0.2n2=27−15n2n^2 = 27 - 15n2n2=27−15n

STEP 3

Move all terms to one side of the equation to set it equal to zero.2n2+15n−27=02n^2 + 15n - 27 = 02n2+15n−27=0

STEP 4

Now that we have the quadratic equation in standard form, we can attempt to factor it. If factoring is not possible or not easily seen, we can use the quadratic formula.

STEP 5

Check if the quadratic can be factored by finding two numbers that multiply to 2×(−27)=−542 \times (-27) = -542×(−27)=−54 and add up to 151515.

STEP 6

After checking for possible factors, we see that there are no two integers that satisfy both conditions. Therefore, we will use the quadratic formula to solve for nnn.

STEP 7

Recall the quadratic formula:n=−b±b2−4ac2an = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}n=2a−b±b2−4ac​​

STEP 8

Identify the coefficients aaa, bbb, and ccc from the quadratic equation 2n2+15n−27=02n^2 + 15n - 27 = 02n2+15n−27=0.a=2,b=15,c=−27a = 2, \quad b = 15, \quad c = -27a=2,b=15,c=−27

STEP 9

Substitute the coefficients into the quadratic formula.n=−15±152−4⋅2⋅(−27)2⋅2n = \frac{-15 \pm \sqrt{15^2 - 4 \cdot 2 \cdot (-27)}}{2 \cdot 2}n=2⋅2−15±152−4⋅2⋅(−27)​​

STEP 10

Calculate the discriminant (b2−4acb^2 - 4acb2−4ac).152−4⋅2⋅(−27)=225+216=44115^2 - 4 \cdot 2 \cdot (-27) = 225 + 216 = 441152−4⋅2⋅(−27)=225+216=441

STEP 11

Take the square root of the discriminant.441=21\sqrt{441} = 21441​=21

STEP 12

Substitute the square root of the discriminant back into the quadratic formula.n=−15±214n = \frac{-15 \pm 21}{4}n=4−15±21​

STEP 13

Solve for the two possible values of nnn.n1=−15+214=64=32n_1 = \frac{-15 + 21}{4} = \frac{6}{4} = \frac{3}{2}n1​=4−15+21​=46​=23​n2=−15−214=−364=−9n_2 = \frac{-15 - 21}{4} = \frac{-36}{4} = -9n2​=4−15−21​=4−36​=−9

STEP 14

The two solutions for nnn are n1=32n_1 = \frac{3}{2}n1​=23​ and n2=−9n_2 = -9n2​=−9.The solutions to the equation 2n2=27−15n2n^2 = 27 - 15n2n2=27−15n are n=32n = \frac{3}{2}n=23​ and n=−9n = -9n=−9.

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Solve the quadratic equation $2n^2 = 27 - 15n$ for the value of $n$ .

Assumptions
1. The equation to solve is $2n^2 = 27 - 15n$ .
2. We will solve for $n$ by rearranging the equation into standard quadratic form and then using the quadratic formula or factoring, if possible.

First, we need to rearrange the equation into the standard quadratic form, which is $ax^2 + bx + c = 0$ .
$2n^2 = 27 - 15n$

Move all terms to one side of the equation to set it equal to zero.
$2n^2 + 15n - 27 = 0$

Check if the quadratic can be factored by finding two numbers that multiply to $2 \times (-27) = -54$ and add up to $15$ .

After checking for possible factors, we see that there are no two integers that satisfy both conditions. Therefore, we will use the quadratic formula to solve for $n$ .

Recall the quadratic formula:
$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Identify the coefficients $a$ , $b$ , and $c$ from the quadratic equation $2n^2 + 15n - 27 = 0$ .
$a = 2, \quad b = 15, \quad c = -27$

Substitute the coefficients into the quadratic formula.
$n = \frac{-15 \pm \sqrt{15^2 - 4 \cdot 2 \cdot (-27)}}{2 \cdot 2}$

Calculate the discriminant ( $b^2 - 4ac$ ).
$15^2 - 4 \cdot 2 \cdot (-27) = 225 + 216 = 441$

Take the square root of the discriminant.
$\sqrt{441} = 21$

Substitute the square root of the discriminant back into the quadratic formula.
$n = \frac{-15 \pm 21}{4}$

Solve for the two possible values of $n$ .
$n_1 = \frac{-15 + 21}{4} = \frac{6}{4} = \frac{3}{2}$
$n_2 = \frac{-15 - 21}{4} = \frac{-36}{4} = -9$

The two solutions for $n$ are $n_1 = \frac{3}{2}$ and $n_2 = -9$ .
The solutions to the equation $2n^2 = 27 - 15n$ are $n = \frac{3}{2}$ and $n = -9$ .