Precalculus Examples

$2 x^{2} + 16 x - 18 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 2$ , $b = 16$ , and $c = - 18$ into the quadratic formula and solve for $x$ .

$\frac{- 16 \pm \sqrt{16^{2} - 4 \cdot (2 \cdot - 18)}}{2 \cdot 2}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $16$ to the power of $2$ .

$x = \frac{- 16 \pm \sqrt{256 - 4 \cdot 2 \cdot - 18}}{2 \cdot 2}$

Step 3.1.2

Multiply $- 4 \cdot 2 \cdot - 18$ .

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Step 3.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{- 16 \pm \sqrt{256 - 8 \cdot - 18}}{2 \cdot 2}$

Step 3.1.2.2

Multiply $- 8$ by $- 18$ .

$x = \frac{- 16 \pm \sqrt{256 + 144}}{2 \cdot 2}$

Step 3.1.3

Add $256$ and $144$ .

$x = \frac{- 16 \pm \sqrt{400}}{2 \cdot 2}$

Step 3.1.4

Rewrite $400$ as $20^{2}$ .

$x = \frac{- 16 \pm \sqrt{20^{2}}}{2 \cdot 2}$

Step 3.1.5

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 16 \pm 20}{2 \cdot 2}$

Step 3.2

Multiply $2$ by $2$ .

$x = \frac{- 16 \pm 20}{4}$

Step 3.3

Simplify $\frac{- 16 \pm 20}{4}$ .

$x = - 4 \pm 5$

Step 4

The final answer is the combination of both solutions.

$x = 1, - 9$