Solve Using the Quadratic Formula 3x^2+16x+5=0

$3 x^{2} + 16 x + 5 = 0$

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 $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{- 16 \pm \sqrt{16^{2} - 4 \cdot (3 \cdot 5)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 16 \pm \sqrt{256 - 4 \cdot 3 \cdot 5}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 5$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 16 \pm \sqrt{256 - 12 \cdot 5}}{2 \cdot 3}$

Multiply $- 12$ by $5$ .

$x = \frac{- 16 \pm \sqrt{256 - 60}}{2 \cdot 3}$

Subtract $60$ from $256$ .

$x = \frac{- 16 \pm \sqrt{196}}{2 \cdot 3}$

Rewrite $196$ as $14^{2}$ .

$x = \frac{- 16 \pm \sqrt{14^{2}}}{2 \cdot 3}$

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

$x = \frac{- 16 \pm 14}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 16 \pm 14}{6}$

Simplify $\frac{- 16 \pm 14}{6}$ .

$x = \frac{- 8 \pm 7}{3}$

Step 4

The final answer is the combination of both solutions.

$x = - \frac{1}{3}, - 5$