Solve left(-xright)^2-4x+5=0

$${ \left(-x \right) }^{ 2 } -4x+5 = 0$$

Solve for x (complex solution)

$$x=2+i$$

$$x=2-i$$

Steps Using the Quadratic Formula

Calculate -x to the power of 2 and get x^{2}.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-\left(-4\right)±\sqrt{16-20}}{2}$$

Add 16 to -20.

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

Take the square root of -4.

$$x=\frac{-\left(-4\right)±2i}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±2i}{2}$$

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

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

Divide 4+2i by 2.

$$x=2+i$$

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

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

Divide 4-2i by 2.

$$x=2-i$$

The equation is now solved.

$$x=2+i x=2-i$$

Steps for Completing the Square

Calculate -x to the power of 2 and get x^{2}.

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

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

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

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.

$$x^{2}-4x+\left(-2\right)^{2}=-5+\left(-2\right)^{2}$$

Square -2.

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

Add -5 to 4.

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

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}.

$$\left(x-2\right)^{2}=-1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{-1}$$

Simplify.

$$x-2=i x-2=-i$$

Add 2 to both sides of the equation.

$$x=2+i x=2-i$$