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

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

Step 1

Subtract $4 x$ from both sides of the equation.

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

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 $c = 2$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot - 1 \cdot 2$ .

Multiply $- 4$ by $- 1$ .

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

Multiply $4$ by $2$ .

$x = \frac{4 \pm \sqrt{16 + 8}}{2 \cdot - 1}$

Add $16$ and $8$ .

$x = \frac{4 \pm \sqrt{24}}{2 \cdot - 1}$

Rewrite $24$ as $2^{2} \cdot 6$ .

Factor $4$ out of $24$ .

$x = \frac{4 \pm \sqrt{4 (6)}}{2 \cdot - 1}$

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

$x = \frac{4 \pm \sqrt{2^{2} \cdot 6}}{2 \cdot - 1}$

Pull terms out from under the radical.

$x = \frac{4 \pm 2 \sqrt{6}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

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

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

$x = \frac{2 \pm \sqrt{6}}{- 1}$

Move the negative one from the denominator of $\frac{2 \pm \sqrt{6}}{- 1}$ .

$x = - 1 \cdot (2 \pm \sqrt{6})$

Rewrite $- 1 \cdot (2 \pm \sqrt{6})$ as $- (2 \pm \sqrt{6})$ .

$x = - (2 \pm \sqrt{6})$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - (2 \pm \sqrt{6})$

Decimal Form:

$x = - 4.44948974 \dots, 0.44948974 \dots$