Solve for q q^2=2q+4

$q^{2} = 2 q + 4$

Subtract $2 q$ from both sides of the equation.

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

Subtract $4$ from both sides of the equation.

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 4$ into the quadratic formula and solve for $q$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 4$ .

$q = \frac{2 \pm \sqrt{4 + 16}}{2 \cdot 1}$

Add $4$ and $16$ .

$q = \frac{2 \pm \sqrt{20}}{2 \cdot 1}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$q = \frac{2 \pm \sqrt{4 (5)}}{2 \cdot 1}$

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

$q = \frac{2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$q = \frac{2 \pm 2 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$q = \frac{2 \pm 2 \sqrt{5}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{5}}{2}$ .

$q = 1 \pm \sqrt{5}$

The final answer is the combination of both solutions.

$q = 1 + \sqrt{5}, 1 - \sqrt{5}$

The result can be shown in multiple forms.

Exact Form:

$q = 1 + \sqrt{5}, 1 - \sqrt{5}$

Decimal Form:

$q = 3.23606797 \dots, - 1.23606797 \dots$