Solve Using the Quadratic Formula 8x^2+18x-5=0

$8 x^{2} + 18 x - 5 = 0$

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 $a = 8$ , $b = 18$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{- 18 \pm \sqrt{18^{2} - 4 \cdot (8 \cdot - 5)}}{2 \cdot 8}$

Step 3

Simplify.

Simplify the numerator.

Raise $18$ to the power of $2$ .

$x = \frac{- 18 \pm \sqrt{324 - 4 \cdot 8 \cdot - 5}}{2 \cdot 8}$

Multiply $- 4 \cdot 8 \cdot - 5$ .

Multiply $- 4$ by $8$ .

$x = \frac{- 18 \pm \sqrt{324 - 32 \cdot - 5}}{2 \cdot 8}$

Multiply $- 32$ by $- 5$ .

$x = \frac{- 18 \pm \sqrt{324 + 160}}{2 \cdot 8}$

Add $324$ and $160$ .

$x = \frac{- 18 \pm \sqrt{484}}{2 \cdot 8}$

Rewrite $484$ as $22^{2}$ .

$x = \frac{- 18 \pm \sqrt{22^{2}}}{2 \cdot 8}$

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

$x = \frac{- 18 \pm 22}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{- 18 \pm 22}{16}$

Simplify $\frac{- 18 \pm 22}{16}$ .

$x = \frac{- 9 \pm 11}{8}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{1}{4}, - \frac{5}{2}$