Solve Using the Quadratic Formula 3x^2-7x-40=0

$3 x^{2} - 7 x - 40 = 0$

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 $a = 3$ , $b = - 7$ , and $c = - 40$ into the quadratic formula and solve for $x$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (3 \cdot - 40)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 7$ to the power of $2$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot 3 \cdot - 40}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 40$ .

Multiply $- 4$ by $3$ .

$x = \frac{7 \pm \sqrt{49 - 12 \cdot - 40}}{2 \cdot 3}$

Multiply $- 12$ by $- 40$ .

$x = \frac{7 \pm \sqrt{49 + 480}}{2 \cdot 3}$

Add $49$ and $480$ .

$x = \frac{7 \pm \sqrt{529}}{2 \cdot 3}$

Rewrite $529$ as $23^{2}$ .

$x = \frac{7 \pm \sqrt{23^{2}}}{2 \cdot 3}$

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

$x = \frac{7 \pm 23}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{7 \pm 23}{6}$

Step 4

The final answer is the combination of both solutions.

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