Math / Algebra

QuestionSolve using the quadratic formula. $-2 t^{2}+9 t+7=0$
Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. $t=$ $\square$ or $t=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the values of $t$ that make this equation true: $-2t^2 + 9t + 7 = 0$ . Watch out! Remember, the quadratic formula has a plus-or-minus sign, so we'll get two answers!
Also, be careful with those negative signs when plugging into the formula.

STEP 2

1. Set up the Quadratic Formula
2. Plug in the values
3. Simplify the Expression
4. Calculate the two solutions

STEP 3

Alright, let's kick things off with the star of the show, the quadratic formula!
It helps us solve equations that look like $ax^2 + bx + c = 0$ .
The formula is:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

STEP 4

Remember, $a$ , $b$ , and $c$ are just the numbers in front of the $x^2$ , $x$ , and the constant term, respectively.

STEP 5

In our equation, $-2t^2 + 9t + 7 = 0$ , we can see that $a = -2$ , $b = 9$ , and $c = 7$ .
Let's plug these bad boys into the quadratic formula!
$t = \frac{-9 \pm \sqrt{9^2 - 4 \cdot (-2) \cdot 7}}{2 \cdot (-2)}$

STEP 6

Now, let's clean this up!
First, we'll tackle the inside of the square root: $9^2$ is $81$ , and $4 \cdot (-2) \cdot 7$ is $-56$ .
So, inside the square root, we have $81 - (-56)$ , which is the same as $81 + 56 = 137$ .
$t = \frac{-9 \pm \sqrt{137}}{-4}$

STEP 7

We can rewrite this as two separate solutions:
$t = \frac{-9 + \sqrt{137}}{-4} \quad \text{or} \quad t = \frac{-9 - \sqrt{137}}{-4}$

STEP 8

Let's multiply the numerator and denominator by $-1$ to make the denominator positive.
Remember, multiplying by $\frac{-1}{-1}$ is the same as multiplying by $1$ , so it doesn't change the value of the expression.
$t = \frac{9 - \sqrt{137}}{4} \quad \text{or} \quad t = \frac{9 + \sqrt{137}}{4}$

STEP 9

Since $\sqrt{137} \approx 11.70$ , we can approximate our solutions:
$t \approx \frac{9 - 11.70}{4} \quad \text{or} \quad t \approx \frac{9 + 11.70}{4}$ $t \approx \frac{-2.70}{4} \quad \text{or} \quad t \approx \frac{20.70}{4}$ $t \approx -0.68 \quad \text{or} \quad t \approx 5.18$

STEP 10

$t \approx -0.68$ or $t \approx 5.18$

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QuestionSolve using the quadratic formula. $-2 t^{2}+9 t+7=0$
Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. $t=$ $\square$ or $t=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the values of $t$ that make this equation true: $-2t^2 + 9t + 7 = 0$ . Watch out! Remember, the quadratic formula has a plus-or-minus sign, so we'll get two answers!
Also, be careful with those negative signs when plugging into the formula.

STEP 2

1. Set up the Quadratic Formula
2. Plug in the values
3. Simplify the Expression
4. Calculate the two solutions

STEP 3

Alright, let's kick things off with the star of the show, the quadratic formula!
It helps us solve equations that look like $ax^2 + bx + c = 0$ .
The formula is:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

STEP 4

Remember, $a$ , $b$ , and $c$ are just the numbers in front of the $x^2$ , $x$ , and the constant term, respectively.

STEP 5

In our equation, $-2t^2 + 9t + 7 = 0$ , we can see that $a = -2$ , $b = 9$ , and $c = 7$ .
Let's plug these bad boys into the quadratic formula!
$t = \frac{-9 \pm \sqrt{9^2 - 4 \cdot (-2) \cdot 7}}{2 \cdot (-2)}$

STEP 6

Now, let's clean this up!
First, we'll tackle the inside of the square root: $9^2$ is $81$ , and $4 \cdot (-2) \cdot 7$ is $-56$ .
So, inside the square root, we have $81 - (-56)$ , which is the same as $81 + 56 = 137$ .
$t = \frac{-9 \pm \sqrt{137}}{-4}$

STEP 7

We can rewrite this as two separate solutions:
$t = \frac{-9 + \sqrt{137}}{-4} \quad \text{or} \quad t = \frac{-9 - \sqrt{137}}{-4}$

STEP 8

Let's multiply the numerator and denominator by $-1$ to make the denominator positive.
Remember, multiplying by $\frac{-1}{-1}$ is the same as multiplying by $1$ , so it doesn't change the value of the expression.
$t = \frac{9 - \sqrt{137}}{4} \quad \text{or} \quad t = \frac{9 + \sqrt{137}}{4}$

STEP 9

Since $\sqrt{137} \approx 11.70$ , we can approximate our solutions:
$t \approx \frac{9 - 11.70}{4} \quad \text{or} \quad t \approx \frac{9 + 11.70}{4}$ $t \approx \frac{-2.70}{4} \quad \text{or} \quad t \approx \frac{20.70}{4}$ $t \approx -0.68 \quad \text{or} \quad t \approx 5.18$

STEP 10

$t \approx -0.68$ or $t \approx 5.18$

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QuestionSolve using the quadratic formula. −2t2+9t+7=0-2 t^{2}+9 t+7=0−2t2+9t+7=0Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. t=t=t= □\square□ or t=t=t= □\square□

Studdy Solution

STEP 1

What is this asking? We need to find the values of ttt that make this equation true: −2t2+9t+7=0-2t^2 + 9t + 7 = 0−2t2+9t+7=0. Watch out! Remember, the quadratic formula has a plus-or-minus sign, so we'll get two answers!Also, be careful with those negative signs when plugging into the formula.

STEP 2

1. Set up the Quadratic Formula2. Plug in the values3. Simplify the Expression4. Calculate the two solutions

STEP 3

Alright, let's **kick things off** with the **star of the show**, the **quadratic formula**!It helps us solve equations that look like ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0.The formula is:x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4ac​​

STEP 4

Remember, aaa, bbb, and ccc are just the numbers in front of the x2x^2x2, xxx, and the constant term, respectively.

STEP 5

STEP 6

STEP 7

We can rewrite this as two separate solutions:t=−9+137−4ort=−9−137−4t = \frac{-9 + \sqrt{137}}{-4} \quad \text{or} \quad t = \frac{-9 - \sqrt{137}}{-4}t=−4−9+137​​ort=−4−9−137​​

STEP 8

STEP 9

STEP 10

t≈−0.68t \approx -0.68t≈−0.68 or t≈5.18t \approx 5.18t≈5.18

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionSolve using the quadratic formula. $-2 t^{2}+9 t+7=0$
Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. $t=$ $\square$ or $t=$ $\square$

What is this asking? We need to find the values of $t$ that make this equation true: $-2t^2 + 9t + 7 = 0$ . Watch out! Remember, the quadratic formula has a plus-or-minus sign, so we'll get two answers!
Also, be careful with those negative signs when plugging into the formula.

1. Set up the Quadratic Formula
2. Plug in the values
3. Simplify the Expression
4. Calculate the two solutions

Alright, let's kick things off with the star of the show, the quadratic formula!
It helps us solve equations that look like $ax^2 + bx + c = 0$ .
The formula is:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Remember, $a$ , $b$ , and $c$ are just the numbers in front of the $x^2$ , $x$ , and the constant term, respectively.

We can rewrite this as two separate solutions:
$t = \frac{-9 + \sqrt{137}}{-4} \quad \text{or} \quad t = \frac{-9 - \sqrt{137}}{-4}$

$t \approx -0.68$ or $t \approx 5.18$