Solve for q 2q^2+6q-5=0

$2 q^{2} + 6 q - 5 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 2$ , $b = 6$ , and $c = - 5$ into the quadratic formula and solve for $q$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (2 \cdot - 5)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$q = \frac{- 6 \pm \sqrt{36 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$q = \frac{- 6 \pm \sqrt{36 - 8 \cdot - 5}}{2 \cdot 2}$

Multiply $- 8$ by $- 5$ .

$q = \frac{- 6 \pm \sqrt{36 + 40}}{2 \cdot 2}$

Add $36$ and $40$ .

$q = \frac{- 6 \pm \sqrt{76}}{2 \cdot 2}$

Rewrite $76$ as $2^{2} \cdot 19$ .

Factor $4$ out of $76$ .

$q = \frac{- 6 \pm \sqrt{4 (19)}}{2 \cdot 2}$

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

$q = \frac{- 6 \pm \sqrt{2^{2} \cdot 19}}{2 \cdot 2}$

Pull terms out from under the radical.

$q = \frac{- 6 \pm 2 \sqrt{19}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$q = \frac{- 6 \pm 2 \sqrt{19}}{4}$

Simplify $\frac{- 6 \pm 2 \sqrt{19}}{4}$ .

$q = \frac{- 3 \pm \sqrt{19}}{2}$

The final answer is the combination of both solutions.

$q = - \frac{3 - \sqrt{19}}{2}, - \frac{3 + \sqrt{19}}{2}$

The result can be shown in multiple forms.

Exact Form:

$q = - \frac{3 - \sqrt{19}}{2}, - \frac{3 + \sqrt{19}}{2}$

Decimal Form:

$q = 0.67944947 \dots, - 3.67944947 \dots$