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

$$( 2 x + 2 ) ^ { 2 } = x ( x + 3 )$$

Solve for x (complex solution)

$$x=\frac{-5+\sqrt{23}i}{6}\approx -0.833333333+0.799305254i$$

$$x=\frac{-\sqrt{23}i-5}{6}\approx -0.833333333-0.799305254i$$

Steps Using the Quadratic Formula

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

$$4x^{2}+8x+4=x\left(x+3\right)$$

Use the distributive property to multiply x by x+3.

$$4x^{2}+8x+4=x^{2}+3x$$

Subtract x^{2} from both sides.

$$4x^{2}+8x+4-x^{2}=3x$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+8x+4=3x$$

Subtract 3x from both sides.

$$3x^{2}+8x+4-3x=0$$

Combine 8x and -3x to get 5x.

$$3x^{2}+5x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 5 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 3\times 4}}{2\times 3}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 3\times 4}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-5±\sqrt{25-12\times 4}}{2\times 3}$$

Multiply -12 times 4.

$$x=\frac{-5±\sqrt{25-48}}{2\times 3}$$

Add 25 to -48.

$$x=\frac{-5±\sqrt{-23}}{2\times 3}$$

Take the square root of -23.

$$x=\frac{-5±\sqrt{23}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-5±\sqrt{23}i}{6}$$

Now solve the equation x=\frac{-5±\sqrt{23}i}{6} when ± is plus. Add -5 to i\sqrt{23}.

$$x=\frac{-5+\sqrt{23}i}{6}$$

Now solve the equation x=\frac{-5±\sqrt{23}i}{6} when ± is minus. Subtract i\sqrt{23} from -5.

$$x=\frac{-\sqrt{23}i-5}{6}$$

The equation is now solved.

$$x=\frac{-5+\sqrt{23}i}{6} x=\frac{-\sqrt{23}i-5}{6}$$

Steps for Completing the Square

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

$$4x^{2}+8x+4=x\left(x+3\right)$$

Use the distributive property to multiply x by x+3.

$$4x^{2}+8x+4=x^{2}+3x$$

Subtract x^{2} from both sides.

$$4x^{2}+8x+4-x^{2}=3x$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+8x+4=3x$$

Subtract 3x from both sides.

$$3x^{2}+8x+4-3x=0$$

Combine 8x and -3x to get 5x.

$$3x^{2}+5x+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}+5x=-4$$

Divide both sides by 3.

$$\frac{3x^{2}+5x}{3}=-\frac{4}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{5}{3}x=-\frac{4}{3}$$

Divide \frac{5}{3}, the coefficient of the x term, by 2 to get \frac{5}{6}. Then add the square of \frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{3}x+\left(\frac{5}{6}\right)^{2}=-\frac{4}{3}+\left(\frac{5}{6}\right)^{2}$$

Square \frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=-\frac{4}{3}+\frac{25}{36}$$

Add -\frac{4}{3} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=-\frac{23}{36}$$

Factor x^{2}+\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{6}\right)^{2}=-\frac{23}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{-\frac{23}{36}}$$

Simplify.

$$x+\frac{5}{6}=\frac{\sqrt{23}i}{6} x+\frac{5}{6}=-\frac{\sqrt{23}i}{6}$$

Subtract \frac{5}{6} from both sides of the equation.

$$x=\frac{-5+\sqrt{23}i}{6} x=\frac{-\sqrt{23}i-5}{6}$$