Solve Using the Quadratic Formula 5x^2+23x-10=0

$5 x^{2} + 23 x - 10 = 0$

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 $a = 5$ , $b = 23$ , and $c = - 10$ into the quadratic formula and solve for $x$ .

$\frac{- 23 \pm \sqrt{23^{2} - 4 \cdot (5 \cdot - 10)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

Raise $23$ to the power of $2$ .

$x = \frac{- 23 \pm \sqrt{529 - 4 \cdot 5 \cdot - 10}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot - 10$ .

Multiply $- 4$ by $5$ .

$x = \frac{- 23 \pm \sqrt{529 - 20 \cdot - 10}}{2 \cdot 5}$

Multiply $- 20$ by $- 10$ .

$x = \frac{- 23 \pm \sqrt{529 + 200}}{2 \cdot 5}$

Add $529$ and $200$ .

$x = \frac{- 23 \pm \sqrt{729}}{2 \cdot 5}$

Rewrite $729$ as $27^{2}$ .

$x = \frac{- 23 \pm \sqrt{27^{2}}}{2 \cdot 5}$

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

$x = \frac{- 23 \pm 27}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 23 \pm 27}{10}$

Step 4

The final answer is the combination of both solutions.

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