Solve Using the Quadratic Formula 12x^2+11x-15=0

$12 x^{2} + 11 x - 15 = 0$

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 $a = 12$ , $b = 11$ , and $c = - 15$ into the quadratic formula and solve for $x$ .

$\frac{- 11 \pm \sqrt{11^{2} - 4 \cdot (12 \cdot - 15)}}{2 \cdot 12}$

Step 3

Simplify.

Simplify the numerator.

Raise $11$ to the power of $2$ .

$x = \frac{- 11 \pm \sqrt{121 - 4 \cdot 12 \cdot - 15}}{2 \cdot 12}$

Multiply $- 4 \cdot 12 \cdot - 15$ .

Multiply $- 4$ by $12$ .

$x = \frac{- 11 \pm \sqrt{121 - 48 \cdot - 15}}{2 \cdot 12}$

Multiply $- 48$ by $- 15$ .

$x = \frac{- 11 \pm \sqrt{121 + 720}}{2 \cdot 12}$

Add $121$ and $720$ .

$x = \frac{- 11 \pm \sqrt{841}}{2 \cdot 12}$

Rewrite $841$ as $29^{2}$ .

$x = \frac{- 11 \pm \sqrt{29^{2}}}{2 \cdot 12}$

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

$x = \frac{- 11 \pm 29}{2 \cdot 12}$

Multiply $2$ by $12$ .

$x = \frac{- 11 \pm 29}{24}$

Step 4

The final answer is the combination of both solutions.

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