Solve (2x+(x-4)^2)=0

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

Solve for x (complex solution)

$$x=3+\sqrt{7}i\approx 3+2.645751311i$$

$$x=-\sqrt{7}i+3\approx 3-2.645751311i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$2x+x^{2}-8x+16=0$$

Combine 2x and -8x to get -6x.

$$-6x+x^{2}+16=0$$

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.

$$x^{2}-6x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 16}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 16}}{2}$$

Multiply -4 times 16.

$$x=\frac{-\left(-6\right)±\sqrt{36-64}}{2}$$

Add 36 to -64.

$$x=\frac{-\left(-6\right)±\sqrt{-28}}{2}$$

Take the square root of -28.

$$x=\frac{-\left(-6\right)±2\sqrt{7}i}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{7}i}{2}$$

Now solve the equation x=\frac{6±2\sqrt{7}i}{2} when ± is plus. Add 6 to 2i\sqrt{7}.

$$x=\frac{6+2\sqrt{7}i}{2}$$

Divide 6+2i\sqrt{7} by 2.

$$x=3+\sqrt{7}i$$

Now solve the equation x=\frac{6±2\sqrt{7}i}{2} when ± is minus. Subtract 2i\sqrt{7} from 6.

$$x=\frac{-2\sqrt{7}i+6}{2}$$

Divide 6-2i\sqrt{7} by 2.

$$x=-\sqrt{7}i+3$$

The equation is now solved.

$$x=3+\sqrt{7}i x=-\sqrt{7}i+3$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$2x+x^{2}-8x+16=0$$

Combine 2x and -8x to get -6x.

$$-6x+x^{2}+16=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$-6x+x^{2}=-16$$

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.

$$x^{2}-6x=-16$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-16+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-16+9$$

Add -16 to 9.

$$x^{2}-6x+9=-7$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=-7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{-7}$$

Simplify.

$$x-3=\sqrt{7}i x-3=-\sqrt{7}i$$

Add 3 to both sides of the equation.

$$x=3+\sqrt{7}i x=-\sqrt{7}i+3$$