Solve 0=x^2-4x+4+5

$$0 = x ^ { 2 } - 4 x + 4 + 5$$

Solve for x (complex solution)

$$x=2+\sqrt{5}i\approx 2+2.236067977i$$

$$x=-\sqrt{5}i+2\approx 2-2.236067977i$$

Steps Using the Quadratic Formula

Add 4 and 5 to get 9.

$$0=x^{2}-4x+9$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-4x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 9}}{2}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 9}}{2}$$

Multiply -4 times 9.

$$x=\frac{-\left(-4\right)±\sqrt{16-36}}{2}$$

Add 16 to -36.

$$x=\frac{-\left(-4\right)±\sqrt{-20}}{2}$$

Take the square root of -20.

$$x=\frac{-\left(-4\right)±2\sqrt{5}i}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{5}i}{2}$$

Now solve the equation x=\frac{4±2\sqrt{5}i}{2} when ± is plus. Add 4 to 2i\sqrt{5}.

$$x=\frac{4+2\sqrt{5}i}{2}$$

Divide 4+2i\sqrt{5} by 2.

$$x=2+\sqrt{5}i$$

Now solve the equation x=\frac{4±2\sqrt{5}i}{2} when ± is minus. Subtract 2i\sqrt{5} from 4.

$$x=\frac{-2\sqrt{5}i+4}{2}$$

Divide 4-2i\sqrt{5} by 2.

$$x=-\sqrt{5}i+2$$

The equation is now solved.

$$x=2+\sqrt{5}i x=-\sqrt{5}i+2$$

Steps for Completing the Square

Add 4 and 5 to get 9.

$$0=x^{2}-4x+9$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-4x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-4x=-9$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=-9+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-9+4$$

Add -9 to 4.

$$x^{2}-4x+4=-5$$

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=-5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{-5}$$

Simplify.

$$x-2=\sqrt{5}i x-2=-\sqrt{5}i$$

Add 2 to both sides of the equation.

$$x=2+\sqrt{5}i x=-\sqrt{5}i+2$$