Solve for q 2q^2+14q-5=0

$2 q^{2} + 14 q - 5 = 0$

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 $a = 2$ , $b = 14$ , and $c = - 5$ into the quadratic formula and solve for $q$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 2 \cdot - 5$ .

Multiply $- 4$ by $2$ .

$q = \frac{- 14 \pm \sqrt{196 - 8 \cdot - 5}}{2 \cdot 2}$

Multiply $- 8$ by $- 5$ .

$q = \frac{- 14 \pm \sqrt{196 + 40}}{2 \cdot 2}$

Add $196$ and $40$ .

$q = \frac{- 14 \pm \sqrt{236}}{2 \cdot 2}$

Rewrite $236$ as $2^{2} \cdot 59$ .

Factor $4$ out of $236$ .

$q = \frac{- 14 \pm \sqrt{4 (59)}}{2 \cdot 2}$

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

$q = \frac{- 14 \pm \sqrt{2^{2} \cdot 59}}{2 \cdot 2}$

Pull terms out from under the radical.

$q = \frac{- 14 \pm 2 \sqrt{59}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$q = \frac{- 14 \pm 2 \sqrt{59}}{4}$

Simplify $\frac{- 14 \pm 2 \sqrt{59}}{4}$ .

$q = \frac{- 7 \pm \sqrt{59}}{2}$

Step 4

The final answer is the combination of both solutions.

$q = - \frac{7 - \sqrt{59}}{2}, - \frac{7 + \sqrt{59}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$q = - \frac{7 - \sqrt{59}}{2}, - \frac{7 + \sqrt{59}}{2}$

Decimal Form:

$q = 0.34057287 \dots, - 7.34057287 \dots$