Solve Using the Quadratic Formula x^2=9-4x

$x^{2} = 9 - 4 x$

Step 1

Move all the expressions to the left side of the equation.

Subtract $9$ from both sides of the equation.

$x^{2} - 9 = - 4 x$

Add $4 x$ to both sides of the equation.

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

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 $a = 1$ , $b = 4$ , and $c = - 9$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot - 9)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 9}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 9$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 9}}{2 \cdot 1}$

Multiply $- 4$ by $- 9$ .

$x = \frac{- 4 \pm \sqrt{16 + 36}}{2 \cdot 1}$

Add $16$ and $36$ .

$x = \frac{- 4 \pm \sqrt{52}}{2 \cdot 1}$

Rewrite $52$ as $2^{2} \cdot 13$ .

Factor $4$ out of $52$ .

$x = \frac{- 4 \pm \sqrt{4 (13)}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 13}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{13}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{13}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{13}}{2}$ .

$x = - 2 \pm \sqrt{13}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - 2 \pm \sqrt{13}$

Decimal Form:

$x = 1.60555127 \dots, - 5.60555127 \dots$