Solve Using the Quadratic Formula 3x^2+14x=-11

$3 x^{2} + 14 x = - 11$

Step 1

Add $11$ to both sides of the equation.

$3 x^{2} + 14 x + 11 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 3$ , $b = 14$ , and $c = 11$ into the quadratic formula and solve for $x$ .

$\frac{- 14 \pm \sqrt{14^{2} - 4 \cdot (3 \cdot 11)}}{2 \cdot 3}$

Step 4

Simplify.

Simplify the numerator.

Raise $14$ to the power of $2$ .

$x = \frac{- 14 \pm \sqrt{196 - 4 \cdot 3 \cdot 11}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 11$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 14 \pm \sqrt{196 - 12 \cdot 11}}{2 \cdot 3}$

Multiply $- 12$ by $11$ .

$x = \frac{- 14 \pm \sqrt{196 - 132}}{2 \cdot 3}$

Subtract $132$ from $196$ .

$x = \frac{- 14 \pm \sqrt{64}}{2 \cdot 3}$

Rewrite $64$ as $8^{2}$ .

$x = \frac{- 14 \pm \sqrt{8^{2}}}{2 \cdot 3}$

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

$x = \frac{- 14 \pm 8}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 14 \pm 8}{6}$

Simplify $\frac{- 14 \pm 8}{6}$ .

$x = \frac{- 7 \pm 4}{3}$

Step 5

The final answer is the combination of both solutions.

$x = - 1, - \frac{11}{3}$