Solve Using the Quadratic Formula 2x^2=1-2x

$2 x^{2} = 1 - 2 x$

Step 1

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

Subtract $1$ from both sides of the equation.

$2 x^{2} - 1 = - 2 x$

Add $2 x$ to both sides of the equation.

$2 x^{2} - 1 + 2 x = 0$

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 $a = 2$ , $b = 2$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $2$ .

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

Multiply $- 8$ by $- 1$ .

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

Add $4$ and $8$ .

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

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

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

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

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

Pull terms out from under the radical.

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

Multiply $2$ by $2$ .

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

Simplify $\frac{- 2 \pm 2 \sqrt{3}}{4}$ .

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

Step 5

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.36602540 \dots, - 1.36602540 \dots$