Solve Using the Quadratic Formula 2x^2+9x-5=0

$2 x^{2} + 9 x - 5 = 0$

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 $a = 2$ , $b = 9$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (2 \cdot - 5)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

Raise $9$ to the power of $2$ .

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 5$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 9 \pm \sqrt{81 - 8 \cdot - 5}}{2 \cdot 2}$

Multiply $- 8$ by $- 5$ .

$x = \frac{- 9 \pm \sqrt{81 + 40}}{2 \cdot 2}$

Add $81$ and $40$ .

$x = \frac{- 9 \pm \sqrt{121}}{2 \cdot 2}$

Rewrite $121$ as $11^{2}$ .

$x = \frac{- 9 \pm \sqrt{11^{2}}}{2 \cdot 2}$

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

$x = \frac{- 9 \pm 11}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 9 \pm 11}{4}$

Step 4

The final answer is the combination of both solutions.

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