Solve for c 2c^2+4c=9

$2 c^{2} + 4 c = 9$

Subtract $9$ from both sides of the equation.

$2 c^{2} + 4 c - 9 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = 4$ , and $c = - 9$ into the quadratic formula and solve for $c$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (2 \cdot - 9)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$c = \frac{- 4 \pm \sqrt{16 - 4 \cdot 2 \cdot - 9}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$c = \frac{- 4 \pm \sqrt{16 - 8 \cdot - 9}}{2 \cdot 2}$

Multiply $- 8$ by $- 9$ .

$c = \frac{- 4 \pm \sqrt{16 + 72}}{2 \cdot 2}$

Add $16$ and $72$ .

$c = \frac{- 4 \pm \sqrt{88}}{2 \cdot 2}$

Rewrite $88$ as $2^{2} \cdot 22$ .

Factor $4$ out of $88$ .

$c = \frac{- 4 \pm \sqrt{4 (22)}}{2 \cdot 2}$

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

$c = \frac{- 4 \pm \sqrt{2^{2} \cdot 22}}{2 \cdot 2}$

Pull terms out from under the radical.

$c = \frac{- 4 \pm 2 \sqrt{22}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$c = \frac{- 4 \pm 2 \sqrt{22}}{4}$

Simplify $\frac{- 4 \pm 2 \sqrt{22}}{4}$ .

$c = \frac{- 2 \pm \sqrt{22}}{2}$

The final answer is the combination of both solutions.

$c = - \frac{2 - \sqrt{22}}{2}, - \frac{2 + \sqrt{22}}{2}$

The result can be shown in multiple forms.

Exact Form:

$c = - \frac{2 - \sqrt{22}}{2}, - \frac{2 + \sqrt{22}}{2}$

Decimal Form:

$c = 1.34520787 \dots, - 3.34520787 \dots$