Solve x^2-2(1+i)x+2i=0

$$x ^ { 2 } - 2 ( 1 + i ) x + 2 i = 0$$

Solve for x

$$x=1+i$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Square -2-2i.

$$x=\frac{2+2i±\sqrt{8i-4\times \left(2i\right)}}{2}$$

Multiply -4 times 2i.

$$x=\frac{2+2i±\sqrt{8i-8i}}{2}$$

Add 8i to -8i.

$$x=\frac{2+2i±\sqrt{0}}{2}$$

Take the square root of 0.

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

Divide 2+2i by 2.

$$x=1+i$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

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

$$\left(x+\left(-1-i\right)\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\left(-1-i\right)\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+\left(-1-i\right)=0 x+\left(-1-i\right)=0$$

Add 1+i to both sides of the equation.

$$x=1+i x=1+i$$

The equation is now solved. Solutions are the same.

$$x=1+i$$