Solve Using the Quadratic Formula 3q^2+q-14=0

$3 q^{2} + q - 14 = 0$

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 $a = 3$ , $b = 1$ , and $c = - 14$ into the quadratic formula and solve for $q$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (3 \cdot - 14)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

One to any power is one.

$q = \frac{- 1 \pm \sqrt{1 - 4 \cdot 3 \cdot - 14}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 14$ .

Multiply $- 4$ by $3$ .

$q = \frac{- 1 \pm \sqrt{1 - 12 \cdot - 14}}{2 \cdot 3}$

Multiply $- 12$ by $- 14$ .

$q = \frac{- 1 \pm \sqrt{1 + 168}}{2 \cdot 3}$

Add $1$ and $168$ .

$q = \frac{- 1 \pm \sqrt{169}}{2 \cdot 3}$

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

$q = \frac{- 1 \pm \sqrt{13^{2}}}{2 \cdot 3}$

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

$q = \frac{- 1 \pm 13}{2 \cdot 3}$

Multiply $2$ by $3$ .

$q = \frac{- 1 \pm 13}{6}$

Step 4

The final answer is the combination of both solutions.

$q = 2, - \frac{7}{3}$