Solve for q 4q^2-2q=-65

$4 q^{2} - 2 q = - 65$

Step 1

Add $65$ to both sides of the equation.

$4 q^{2} - 2 q + 65 = 0$

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 $a = 4$ , $b = - 2$ , and $c = 65$ into the quadratic formula and solve for $q$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (4 \cdot 65)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

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

$q = \frac{2 \pm \sqrt{4 - 4 \cdot 4 \cdot 65}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 65$ .

Multiply $- 4$ by $4$ .

$q = \frac{2 \pm \sqrt{4 - 16 \cdot 65}}{2 \cdot 4}$

Multiply $- 16$ by $65$ .

$q = \frac{2 \pm \sqrt{4 - 1040}}{2 \cdot 4}$

Subtract $1040$ from $4$ .

$q = \frac{2 \pm \sqrt{- 1036}}{2 \cdot 4}$

Rewrite $- 1036$ as $- 1 (1036)$ .

$q = \frac{2 \pm \sqrt{- 1 \cdot 1036}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (1036)}$ as $\sqrt{- 1} \cdot \sqrt{1036}$ .

$q = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{1036}}{2 \cdot 4}$

Rewrite $\sqrt{- 1}$ as $i$ .

$q = \frac{2 \pm i \cdot \sqrt{1036}}{2 \cdot 4}$

Rewrite $1036$ as $2^{2} \cdot 259$ .

Factor $4$ out of $1036$ .

$q = \frac{2 \pm i \cdot \sqrt{4 (259)}}{2 \cdot 4}$

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

$q = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 259}}{2 \cdot 4}$

Pull terms out from under the radical.

$q = \frac{2 \pm i \cdot (2 \sqrt{259})}{2 \cdot 4}$

Move $2$ to the left of $i$ .

$q = \frac{2 \pm 2 i \sqrt{259}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$q = \frac{2 \pm 2 i \sqrt{259}}{8}$

Simplify $\frac{2 \pm 2 i \sqrt{259}}{8}$ .

$q = \frac{1 \pm i \sqrt{259}}{4}$

Step 5

The final answer is the combination of both solutions.

$q = \frac{1 + i \sqrt{259}}{4}, \frac{1 - i \sqrt{259}}{4}$