Precalculus Examples

$- 16 x^{2} + 96 x + 15 = h$

Step 1

Subtract $h$ from both sides of the equation.

$- 16 x^{2} + 96 x + 15 - h = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = - 16$ , $b = 96$ , and $c = 15 - h$ into the quadratic formula and solve for $x$ .

$\frac{- 96 \pm \sqrt{96^{2} - 4 \cdot (- 16 \cdot (15 - h))}}{2 \cdot - 16}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $96$ to the power of $2$ .

$x = \frac{- 96 \pm \sqrt{9216 - 4 \cdot - 16 \cdot (15 - h)}}{2 \cdot - 16}$

Step 4.1.2

Multiply $- 4$ by $- 16$ .

$x = \frac{- 96 \pm \sqrt{9216 + 64 \cdot (15 - h)}}{2 \cdot - 16}$

Step 4.1.3

Apply the distributive property.

$x = \frac{- 96 \pm \sqrt{9216 + 64 \cdot 15 + 64 (- h)}}{2 \cdot - 16}$

Step 4.1.4

Multiply $64$ by $15$ .

$x = \frac{- 96 \pm \sqrt{9216 + 960 + 64 (- h)}}{2 \cdot - 16}$

Step 4.1.5

Multiply $- 1$ by $64$ .

$x = \frac{- 96 \pm \sqrt{9216 + 960 - 64 h}}{2 \cdot - 16}$

Step 4.1.6

Add $9216$ and $960$ .

$x = \frac{- 96 \pm \sqrt{10176 - 64 h}}{2 \cdot - 16}$

Step 4.1.7

Factor $64$ out of $10176 - 64 h$ .

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Step 4.1.7.1

Factor $64$ out of $10176$ .

$x = \frac{- 96 \pm \sqrt{64 (159) - 64 h}}{2 \cdot - 16}$

Step 4.1.7.2

Factor $64$ out of $- 64 h$ .

$x = \frac{- 96 \pm \sqrt{64 (159) + 64 (- h)}}{2 \cdot - 16}$

Step 4.1.7.3

Factor $64$ out of $64 (159) + 64 (- h)$ .

$x = \frac{- 96 \pm \sqrt{64 (159 - h)}}{2 \cdot - 16}$

Step 4.1.8

Rewrite $64$ as $8^{2}$ .

$x = \frac{- 96 \pm \sqrt{8^{2} (159 - h)}}{2 \cdot - 16}$

Step 4.1.9

Pull terms out from under the radical.

$x = \frac{- 96 \pm 8 \sqrt{159 - h}}{2 \cdot - 16}$

Step 4.2

Multiply $2$ by $- 16$ .

$x = \frac{- 96 \pm 8 \sqrt{159 - h}}{- 32}$

Step 4.3

Simplify $\frac{- 96 \pm 8 \sqrt{159 - h}}{- 32}$ .

$x = \frac{12 \pm \sqrt{159 - h}}{4}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{12 + \sqrt{159 - h}}{4}$

$x = \frac{12 - \sqrt{159 - h}}{4}$