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

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

Step 1

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

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

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

Add $4$ to both sides of the equation.

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

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 $c = 4$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $4$ .

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

Subtract $16$ from $16$ .

$x = \frac{4 \pm \sqrt{0}}{2 \cdot 1}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{4 \pm \sqrt{0^{2}}}{2 \cdot 1}$

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

$x = \frac{4 \pm 0}{2 \cdot 1}$

$4$ plus or minus $0$ is $4$ .

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

Multiply $2$ by $1$ .

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

Divide $4$ by $2$ .

$x = 2$

Step 5

The final answer is the combination of both solutions.

$x = 2$ Double roots