Solve Using the Quadratic Formula 3x^2+11x-4=0

$3 x^{2} + 11 x - 4 = 0$

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 $a = 3$ , $b = 11$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 11 \pm \sqrt{121 - 4 \cdot 3 \cdot - 4}}{2 \cdot 3}$

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

Multiply $- 4$ by $3$ .

$x = \frac{- 11 \pm \sqrt{121 - 12 \cdot - 4}}{2 \cdot 3}$

Multiply $- 12$ by $- 4$ .

$x = \frac{- 11 \pm \sqrt{121 + 48}}{2 \cdot 3}$

Add $121$ and $48$ .

$x = \frac{- 11 \pm \sqrt{169}}{2 \cdot 3}$

Rewrite $169$ as $13^{2}$ .

$x = \frac{- 11 \pm \sqrt{13^{2}}}{2 \cdot 3}$

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

$x = \frac{- 11 \pm 13}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 11 \pm 13}{6}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{1}{3}, - 4$