Solve Using the Quadratic Formula 12x^2+17x-40=0

$12 x^{2} + 17 x - 40 = 0$

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 $a = 12$ , $b = 17$ , and $c = - 40$ into the quadratic formula and solve for $x$ .

$\frac{- 17 \pm \sqrt{17^{2} - 4 \cdot (12 \cdot - 40)}}{2 \cdot 12}$

Step 3

Simplify.

Simplify the numerator.

Raise $17$ to the power of $2$ .

$x = \frac{- 17 \pm \sqrt{289 - 4 \cdot 12 \cdot - 40}}{2 \cdot 12}$

Multiply $- 4 \cdot 12 \cdot - 40$ .

Multiply $- 4$ by $12$ .

$x = \frac{- 17 \pm \sqrt{289 - 48 \cdot - 40}}{2 \cdot 12}$

Multiply $- 48$ by $- 40$ .

$x = \frac{- 17 \pm \sqrt{289 + 1920}}{2 \cdot 12}$

Add $289$ and $1920$ .

$x = \frac{- 17 \pm \sqrt{2209}}{2 \cdot 12}$

Rewrite $2209$ as $47^{2}$ .

$x = \frac{- 17 \pm \sqrt{47^{2}}}{2 \cdot 12}$

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

$x = \frac{- 17 \pm 47}{2 \cdot 12}$

Multiply $2$ by $12$ .

$x = \frac{- 17 \pm 47}{24}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{5}{4}, - \frac{8}{3}$