Solve 2*(x-2)-2=(x+3)^2

$$2 \times ( x - 2 ) - 2 = ( x + 3 ) ^ { 2 }$$

Solve for x (complex solution)

$$x=-\sqrt{11}i-2\approx -2-3.31662479i$$

$$x=-2+\sqrt{11}i\approx -2+3.31662479i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by x-2.

$$2x-4-2=\left(x+3\right)^{2}$$

Subtract 2 from -4 to get -6.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

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

Subtract x^{2} from both sides.

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

Subtract 6x from both sides.

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

Combine 2x and -6x to get -4x.

$$-4x-6-x^{2}=9$$

Subtract 9 from both sides.

$$-4x-6-x^{2}-9=0$$

Subtract 9 from -6 to get -15.

$$-4x-15-x^{2}=0$$

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.

$$-x^{2}-4x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -4 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\left(-15\right)}}{2\left(-1\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\left(-15\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\left(-15\right)}}{2\left(-1\right)}$$

Multiply 4 times -15.

$$x=\frac{-\left(-4\right)±\sqrt{16-60}}{2\left(-1\right)}$$

Add 16 to -60.

$$x=\frac{-\left(-4\right)±\sqrt{-44}}{2\left(-1\right)}$$

Take the square root of -44.

$$x=\frac{-\left(-4\right)±2\sqrt{11}i}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{11}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±2\sqrt{11}i}{-2}$$

Now solve the equation x=\frac{4±2\sqrt{11}i}{-2} when ± is plus. Add 4 to 2i\sqrt{11}.

$$x=\frac{4+2\sqrt{11}i}{-2}$$

Divide 4+2i\sqrt{11} by -2.

$$x=-\sqrt{11}i-2$$

Now solve the equation x=\frac{4±2\sqrt{11}i}{-2} when ± is minus. Subtract 2i\sqrt{11} from 4.

$$x=\frac{-2\sqrt{11}i+4}{-2}$$

Divide 4-2i\sqrt{11} by -2.

$$x=-2+\sqrt{11}i$$

The equation is now solved.

$$x=-\sqrt{11}i-2 x=-2+\sqrt{11}i$$

Steps for Completing the Square

Use the distributive property to multiply 2 by x-2.

$$2x-4-2=\left(x+3\right)^{2}$$

Subtract 2 from -4 to get -6.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

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

Subtract x^{2} from both sides.

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

Subtract 6x from both sides.

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

Combine 2x and -6x to get -4x.

$$-4x-6-x^{2}=9$$

Add 6 to both sides.

$$-4x-x^{2}=9+6$$

Add 9 and 6 to get 15.

$$-4x-x^{2}=15$$

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.

$$-x^{2}-4x=15$$

Divide both sides by -1.

$$\frac{-x^{2}-4x}{-1}=\frac{15}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{4}{-1}\right)x=\frac{15}{-1}$$

Divide -4 by -1.

$$x^{2}+4x=\frac{15}{-1}$$

Divide 15 by -1.

$$x^{2}+4x=-15$$

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.

$$x^{2}+4x+2^{2}=-15+2^{2}$$

Square 2.

$$x^{2}+4x+4=-15+4$$

Add -15 to 4.

$$x^{2}+4x+4=-11$$

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}.

$$\left(x+2\right)^{2}=-11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{-11}$$

Simplify.

$$x+2=\sqrt{11}i x+2=-\sqrt{11}i$$

Subtract 2 from both sides of the equation.

$$x=-2+\sqrt{11}i x=-\sqrt{11}i-2$$