Solve (2x+1)^2+(2x+3)^2=0

$$( 2 x + 1 ) ^ { 2 } + ( 2 x + 3 ) ^ { 2 } = 0$$

Solve for x (complex solution)

$$x=-1+\frac{1}{2}i=-1+0.5i$$

$$x=-1-\frac{1}{2}i=-1-0.5i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$4x^{2}+4x+1+\left(2x+3\right)^{2}=0$$

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

$$4x^{2}+4x+1+4x^{2}+12x+9=0$$

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$8x^{2}+4x+1+12x+9=0$$

Combine 4x and 12x to get 16x.

$$8x^{2}+16x+1+9=0$$

Add 1 and 9 to get 10.

$$8x^{2}+16x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 16 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\times 8\times 10}}{2\times 8}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 8\times 10}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-16±\sqrt{256-32\times 10}}{2\times 8}$$

Multiply -32 times 10.

$$x=\frac{-16±\sqrt{256-320}}{2\times 8}$$

Add 256 to -320.

$$x=\frac{-16±\sqrt{-64}}{2\times 8}$$

Take the square root of -64.

$$x=\frac{-16±8i}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-16±8i}{16}$$

Now solve the equation x=\frac{-16±8i}{16} when ± is plus. Add -16 to 8i.

$$x=\frac{-16+8i}{16}$$

Divide -16+8i by 16.

$$x=-1+\frac{1}{2}i$$

Now solve the equation x=\frac{-16±8i}{16} when ± is minus. Subtract 8i from -16.

$$x=\frac{-16-8i}{16}$$

Divide -16-8i by 16.

$$x=-1-\frac{1}{2}i$$

The equation is now solved.

$$x=-1+\frac{1}{2}i x=-1-\frac{1}{2}i$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$4x^{2}+4x+1+\left(2x+3\right)^{2}=0$$

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

$$4x^{2}+4x+1+4x^{2}+12x+9=0$$

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$8x^{2}+4x+1+12x+9=0$$

Combine 4x and 12x to get 16x.

$$8x^{2}+16x+1+9=0$$

Add 1 and 9 to get 10.

$$8x^{2}+16x+10=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$8x^{2}+16x=-10$$

Divide both sides by 8.

$$\frac{8x^{2}+16x}{8}=-\frac{10}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{16}{8}x=-\frac{10}{8}$$

Divide 16 by 8.

$$x^{2}+2x=-\frac{10}{8}$$

Reduce the fraction \frac{-10}{8} to lowest terms by extracting and canceling out 2.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=-\frac{5}{4}+1^{2}$$

Square 1.

$$x^{2}+2x+1=-\frac{5}{4}+1$$

Add -\frac{5}{4} to 1.

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

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=\frac{1}{2}i x+1=-\frac{1}{2}i$$

Subtract 1 from both sides of the equation.

$$x=-1+\frac{1}{2}i x=-1-\frac{1}{2}i$$