Solve Using the Quadratic Formula 10x^2+19x-15=0

$10 x^{2} + 19 x - 15 = 0$

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 $a = 10$ , $b = 19$ , and $c = - 15$ into the quadratic formula and solve for $x$ .

$\frac{- 19 \pm \sqrt{19^{2} - 4 \cdot (10 \cdot - 15)}}{2 \cdot 10}$

Step 3

Simplify.

Simplify the numerator.

Raise $19$ to the power of $2$ .

$x = \frac{- 19 \pm \sqrt{361 - 4 \cdot 10 \cdot - 15}}{2 \cdot 10}$

Multiply $- 4 \cdot 10 \cdot - 15$ .

Multiply $- 4$ by $10$ .

$x = \frac{- 19 \pm \sqrt{361 - 40 \cdot - 15}}{2 \cdot 10}$

Multiply $- 40$ by $- 15$ .

$x = \frac{- 19 \pm \sqrt{361 + 600}}{2 \cdot 10}$

Add $361$ and $600$ .

$x = \frac{- 19 \pm \sqrt{961}}{2 \cdot 10}$

Rewrite $961$ as $31^{2}$ .

$x = \frac{- 19 \pm \sqrt{31^{2}}}{2 \cdot 10}$

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

$x = \frac{- 19 \pm 31}{2 \cdot 10}$

Multiply $2$ by $10$ .

$x = \frac{- 19 \pm 31}{20}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{3}{5}, - \frac{5}{2}$