Solve Using the Quadratic Formula 3-x^2=2x

$3 - x^{2} = 2 x$

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

$3 - x^{2} - 2 x = 0$

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 $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (- 1 \cdot 3)}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 1 \cdot 3}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{2 \pm \sqrt{4 + 4 \cdot 3}}{2 \cdot - 1}$

Multiply $4$ by $3$ .

$x = \frac{2 \pm \sqrt{4 + 12}}{2 \cdot - 1}$

Add $4$ and $12$ .

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

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical, assuming positive real numbers.

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

Multiply $2$ by $- 1$ .

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

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

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

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

$x = - 1 \cdot (1 \pm 2)$

Rewrite $- 1 \cdot (1 \pm 2)$ as $- (1 \pm 2)$ .

$x = - (1 \pm 2)$

The final answer is the combination of both solutions.

$x = - 3, 1$