Find All Complex Solutions x^2-4x+4=2x-1-x^2

$x^{2} - 4 x + 4 = 2 x - 1 - x^{2}$

Step 1

Move all terms to the left side of the equation and simplify.

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

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

$x^{2} - 4 x + 4 - 2 x = - 1 - x^{2}$

Add $1$ to both sides of the equation.

$x^{2} - 4 x + 4 - 2 x + 1 = - x^{2}$

Add $x^{2}$ to both sides of the equation.

$x^{2} - 4 x + 4 - 2 x + 1 + x^{2} = 0$

Simplify $x^{2} - 4 x + 4 - 2 x + 1 + x^{2}$ .

Add $x^{2}$ and $x^{2}$ .

$2 x^{2} - 4 x + 4 - 2 x + 1 = 0$

Subtract $2 x$ from $- 4 x$ .

$2 x^{2} - 6 x + 4 + 1 = 0$

Add $4$ and $1$ .

$2 x^{2} - 6 x + 5 = 0$

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 $a = 2$ , $b = - 6$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (2 \cdot 5)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 2 \cdot 5}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 5$ .

Multiply $- 4$ by $2$ .

$x = \frac{6 \pm \sqrt{36 - 8 \cdot 5}}{2 \cdot 2}$

Multiply $- 8$ by $5$ .

$x = \frac{6 \pm \sqrt{36 - 40}}{2 \cdot 2}$

Subtract $40$ from $36$ .

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

Rewrite $- 4$ as $- 1 (4)$ .

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

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

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

Rewrite $\sqrt{- 1}$ as $i$ .

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

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

$x = \frac{6 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 2}$

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

$x = \frac{6 \pm i \cdot 2}{2 \cdot 2}$

Move $2$ to the left of $i$ .

$x = \frac{6 \pm 2 i}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{6 \pm 2 i}{4}$

Simplify $\frac{6 \pm 2 i}{4}$ .

$x = \frac{3 \pm i}{2}$

Find All Complex Solutions x^2-4x+4=2x-1-x^2

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