Find the Roots (Zeros) f(x)=2x^2+16x-9

$f (x) = 2 x^{2} + 16 x - 9$

Step 1

Set $2 x^{2} + 16 x - 9$ equal to $0$ .

$2 x^{2} + 16 x - 9 = 0$

Step 2

Solve for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = 16$ , and $c = - 9$ into the quadratic formula and solve for $x$ .

$\frac{- 16 \pm \sqrt{16^{2} - 4 \cdot (2 \cdot - 9)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 16 \pm \sqrt{256 - 4 \cdot 2 \cdot - 9}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 9$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 16 \pm \sqrt{256 - 8 \cdot - 9}}{2 \cdot 2}$

Multiply $- 8$ by $- 9$ .

$x = \frac{- 16 \pm \sqrt{256 + 72}}{2 \cdot 2}$

Add $256$ and $72$ .

$x = \frac{- 16 \pm \sqrt{328}}{2 \cdot 2}$

Rewrite $328$ as $2^{2} \cdot 82$ .

Factor $4$ out of $328$ .

$x = \frac{- 16 \pm \sqrt{4 (82)}}{2 \cdot 2}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 16 \pm \sqrt{2^{2} \cdot 82}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{- 16 \pm 2 \sqrt{82}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 16 \pm 2 \sqrt{82}}{4}$

Simplify $\frac{- 16 \pm 2 \sqrt{82}}{4}$ .

$x = \frac{- 8 \pm \sqrt{82}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{8 - \sqrt{82}}{2}, - \frac{8 + \sqrt{82}}{2}$

$x = \frac{- 8 \pm \sqrt{82}}{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 8 \pm \sqrt{82}}{2}$

Decimal Form:

$x = 0.52769256 \dots, - 8.52769256 \dots$

Step 4