Solve Using the Quadratic Formula 10x^2-23x+12=0

$10 x^{2} - 23 x + 12 = 0$

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 $a = 10$ , $b = - 23$ , and $c = 12$ into the quadratic formula and solve for $x$ .

$\frac{23 \pm \sqrt{{(- 23)}^{2} - 4 \cdot (10 \cdot 12)}}{2 \cdot 10}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{23 \pm \sqrt{529 - 4 \cdot 10 \cdot 12}}{2 \cdot 10}$

Multiply $- 4 \cdot 10 \cdot 12$ .

Multiply $- 4$ by $10$ .

$x = \frac{23 \pm \sqrt{529 - 40 \cdot 12}}{2 \cdot 10}$

Multiply $- 40$ by $12$ .

$x = \frac{23 \pm \sqrt{529 - 480}}{2 \cdot 10}$

Subtract $480$ from $529$ .

$x = \frac{23 \pm \sqrt{49}}{2 \cdot 10}$

Rewrite $49$ as $7^{2}$ .

$x = \frac{23 \pm \sqrt{7^{2}}}{2 \cdot 10}$

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

$x = \frac{23 \pm 7}{2 \cdot 10}$

Multiply $2$ by $10$ .

$x = \frac{23 \pm 7}{20}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{3}{2}, \frac{4}{5}$