Solve Using the Quadratic Formula 10x^2-31x+15=0

$10 x^{2} - 31 x + 15 = 0$

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 $a = 10$ , $b = - 31$ , and $c = 15$ into the quadratic formula and solve for $x$ .

$\frac{31 \pm \sqrt{{(- 31)}^{2} - 4 \cdot (10 \cdot 15)}}{2 \cdot 10}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 31$ to the power of $2$ .

$x = \frac{31 \pm \sqrt{961 - 4 \cdot 10 \cdot 15}}{2 \cdot 10}$

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

Multiply $- 4$ by $10$ .

$x = \frac{31 \pm \sqrt{961 - 40 \cdot 15}}{2 \cdot 10}$

Multiply $- 40$ by $15$ .

$x = \frac{31 \pm \sqrt{961 - 600}}{2 \cdot 10}$

Subtract $600$ from $961$ .

$x = \frac{31 \pm \sqrt{361}}{2 \cdot 10}$

Rewrite $361$ as $19^{2}$ .

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

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

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

Multiply $2$ by $10$ .

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

Step 4

The final answer is the combination of both solutions.

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