Solved on Dec 12, 2023

Solve the quadratic equation $2y(y+5)-5=0$ using the quadratic formula. Provide the solution(s) in the form $y=\square$ .

STEP 1

Assumptions
1. The equation given is $2y(y+5)-5=0$ .
2. We need to solve for $y$ using the Quadratic Formula.

STEP 2

First, we need to expand the equation and write it in the standard quadratic form, $ay^2 + by + c = 0$ .
$2y(y+5)-5=2y^2+10y-5$

STEP 3

Now, we can identify the coefficients $a$ , $b$ , and $c$ from the quadratic equation.
$a = 2, \quad b = 10, \quad c = -5$

STEP 4

The Quadratic Formula is given by:
$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

STEP 5

Substitute the values of $a$ , $b$ , and $c$ into the Quadratic Formula.
$y = \frac{-10 \pm \sqrt{10^2 - 4(2)(-5)}}{2(2)}$

STEP 6

Simplify the expression under the square root (the discriminant).
$\sqrt{10^2 - 4(2)(-5)} = \sqrt{100 + 40} = \sqrt{140}$

STEP 7

Further simplify the square root by factoring out perfect squares.
$\sqrt{140} = \sqrt{4 \cdot 35} = 2\sqrt{35}$

STEP 8

Now, substitute the simplified discriminant back into the Quadratic Formula.
$y = \frac{-10 \pm 2\sqrt{35}}{4}$

STEP 9

Simplify the fraction by dividing both the numerator and the denominator by the common factor of 2.
$y = \frac{-5 \pm \sqrt{35}}{2}$

STEP 10

We have two solutions for $y$ , one with the plus sign and one with the minus sign.
$y_1 = \frac{-5 + \sqrt{35}}{2}, \quad y_2 = \frac{-5 - \sqrt{35}}{2}$
The solutions to the equation $2y(y+5)-5=0$ are:
$y=\frac{-5 + \sqrt{35}}{2}, \frac{-5 - \sqrt{35}}{2}$

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Solve the quadratic equation $2y(y+5)-5=0$ using the quadratic formula. Provide the solution(s) in the form $y=\square$ .

STEP 1

Assumptions
1. The equation given is $2y(y+5)-5=0$ .
2. We need to solve for $y$ using the Quadratic Formula.

STEP 2

First, we need to expand the equation and write it in the standard quadratic form, $ay^2 + by + c = 0$ .
$2y(y+5)-5=2y^2+10y-5$

STEP 3

Now, we can identify the coefficients $a$ , $b$ , and $c$ from the quadratic equation.
$a = 2, \quad b = 10, \quad c = -5$

STEP 4

The Quadratic Formula is given by:
$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

STEP 5

Substitute the values of $a$ , $b$ , and $c$ into the Quadratic Formula.
$y = \frac{-10 \pm \sqrt{10^2 - 4(2)(-5)}}{2(2)}$

STEP 6

Simplify the expression under the square root (the discriminant).
$\sqrt{10^2 - 4(2)(-5)} = \sqrt{100 + 40} = \sqrt{140}$

STEP 7

Further simplify the square root by factoring out perfect squares.
$\sqrt{140} = \sqrt{4 \cdot 35} = 2\sqrt{35}$

STEP 8

Now, substitute the simplified discriminant back into the Quadratic Formula.
$y = \frac{-10 \pm 2\sqrt{35}}{4}$

STEP 9

Simplify the fraction by dividing both the numerator and the denominator by the common factor of 2.
$y = \frac{-5 \pm \sqrt{35}}{2}$

STEP 10

We have two solutions for $y$ , one with the plus sign and one with the minus sign.
$y_1 = \frac{-5 + \sqrt{35}}{2}, \quad y_2 = \frac{-5 - \sqrt{35}}{2}$
The solutions to the equation $2y(y+5)-5=0$ are:
$y=\frac{-5 + \sqrt{35}}{2}, \frac{-5 - \sqrt{35}}{2}$

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Solve the quadratic equation 2y(y+5)−5=02y(y+5)-5=02y(y+5)−5=0 using the quadratic formula. Provide the solution(s) in the form y=□y=\squarey=□.

STEP 1

Assumptions1. The equation given is 2y(y+5)−5=02y(y+5)-5=02y(y+5)−5=0.2. We need to solve for yyy using the Quadratic Formula.

STEP 2

First, we need to expand the equation and write it in the standard quadratic form, ay2+by+c=0ay^2 + by + c = 0ay2+by+c=0.2y(y+5)−5=2y2+10y−52y(y+5)-5=2y^2+10y-52y(y+5)−5=2y2+10y−5

STEP 3

Now, we can identify the coefficients aaa, bbb, and ccc from the quadratic equation.a=2,b=10,c=−5a = 2, \quad b = 10, \quad c = -5a=2,b=10,c=−5

STEP 4

The Quadratic Formula is given by:y=−b±b2−4ac2ay = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}y=2a−b±b2−4ac​​

STEP 5

Substitute the values of aaa, bbb, and ccc into the Quadratic Formula.y=−10±102−4(2)(−5)2(2)y = \frac{-10 \pm \sqrt{10^2 - 4(2)(-5)}}{2(2)}y=2(2)−10±102−4(2)(−5)​​

STEP 6

Simplify the expression under the square root (the discriminant).102−4(2)(−5)=100+40=140\sqrt{10^2 - 4(2)(-5)} = \sqrt{100 + 40} = \sqrt{140}102−4(2)(−5)​=100+40​=140​

STEP 7

Further simplify the square root by factoring out perfect squares.140=4⋅35=235\sqrt{140} = \sqrt{4 \cdot 35} = 2\sqrt{35}140​=4⋅35​=235​

STEP 8

Now, substitute the simplified discriminant back into the Quadratic Formula.y=−10±2354y = \frac{-10 \pm 2\sqrt{35}}{4}y=4−10±235​​

STEP 9

Simplify the fraction by dividing both the numerator and the denominator by the common factor of 2.y=−5±352y = \frac{-5 \pm \sqrt{35}}{2}y=2−5±35​​

STEP 10

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Solve the quadratic equation $2y(y+5)-5=0$ using the quadratic formula. Provide the solution(s) in the form $y=\square$ .

Assumptions
1. The equation given is $2y(y+5)-5=0$ .
2. We need to solve for $y$ using the Quadratic Formula.

First, we need to expand the equation and write it in the standard quadratic form, $ay^2 + by + c = 0$ .
$2y(y+5)-5=2y^2+10y-5$

Now, we can identify the coefficients $a$ , $b$ , and $c$ from the quadratic equation.
$a = 2, \quad b = 10, \quad c = -5$

The Quadratic Formula is given by:
$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values of $a$ , $b$ , and $c$ into the Quadratic Formula.
$y = \frac{-10 \pm \sqrt{10^2 - 4(2)(-5)}}{2(2)}$

Simplify the expression under the square root (the discriminant).
$\sqrt{10^2 - 4(2)(-5)} = \sqrt{100 + 40} = \sqrt{140}$

Further simplify the square root by factoring out perfect squares.
$\sqrt{140} = \sqrt{4 \cdot 35} = 2\sqrt{35}$

Now, substitute the simplified discriminant back into the Quadratic Formula.
$y = \frac{-10 \pm 2\sqrt{35}}{4}$

Simplify the fraction by dividing both the numerator and the denominator by the common factor of 2.
$y = \frac{-5 \pm \sqrt{35}}{2}$