Solve Using the Quadratic Formula 3y^2+14y=5

$3 y^{2} + 14 y = 5$

Step 1

Subtract $5$ from both sides of the equation.

$3 y^{2} + 14 y - 5 = 0$

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

$\frac{- 14 \pm \sqrt{14^{2} - 4 \cdot (3 \cdot - 5)}}{2 \cdot 3}$

Step 4

Simplify.

Simplify the numerator.

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

$y = \frac{- 14 \pm \sqrt{196 - 4 \cdot 3 \cdot - 5}}{2 \cdot 3}$

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

Multiply $- 4$ by $3$ .

$y = \frac{- 14 \pm \sqrt{196 - 12 \cdot - 5}}{2 \cdot 3}$

Multiply $- 12$ by $- 5$ .

$y = \frac{- 14 \pm \sqrt{196 + 60}}{2 \cdot 3}$

Add $196$ and $60$ .

$y = \frac{- 14 \pm \sqrt{256}}{2 \cdot 3}$

Rewrite $256$ as $16^{2}$ .

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

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

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

Multiply $2$ by $3$ .

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

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

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

Step 5

The final answer is the combination of both solutions.

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