Solve Using the Quadratic Formula 3x^2+16x-3=0

$3 x^{2} + 16 x - 3 = 0$

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 $a = 3$ , $b = 16$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{- 16 \pm \sqrt{16^{2} - 4 \cdot (3 \cdot - 3)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

Raise $16$ to the power of $2$ .

$x = \frac{- 16 \pm \sqrt{256 - 4 \cdot 3 \cdot - 3}}{2 \cdot 3}$

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

Multiply $- 4$ by $3$ .

$x = \frac{- 16 \pm \sqrt{256 - 12 \cdot - 3}}{2 \cdot 3}$

Multiply $- 12$ by $- 3$ .

$x = \frac{- 16 \pm \sqrt{256 + 36}}{2 \cdot 3}$

Add $256$ and $36$ .

$x = \frac{- 16 \pm \sqrt{292}}{2 \cdot 3}$

Rewrite $292$ as $2^{2} \cdot 73$ .

Factor $4$ out of $292$ .

$x = \frac{- 16 \pm \sqrt{4 (73)}}{2 \cdot 3}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 16 \pm \sqrt{2^{2} \cdot 73}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{- 16 \pm 2 \sqrt{73}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 16 \pm 2 \sqrt{73}}{6}$

Simplify $\frac{- 16 \pm 2 \sqrt{73}}{6}$ .

$x = \frac{- 8 \pm \sqrt{73}}{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 8 \pm \sqrt{73}}{3}$

Decimal Form:

$x = 0.18133458 \dots, - 5.51466791 \dots$