Solve left(2x+1right)^2=(2x-1^2

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

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{7}i}{4}\approx -0.25+0.661437828i$$

$$x=\frac{-\sqrt{7}i-1}{4}\approx -0.25-0.661437828i$$

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}.

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

Calculate 1 to the power of 2 and get 1.

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

Subtract 2x from both sides.

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

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

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

Add 1 to both sides.

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

Add 1 and 1 to get 2.

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

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

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

Square 2.

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

Multiply -4 times 4.

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

Multiply -16 times 2.

$$x=\frac{-2±\sqrt{4-32}}{2\times 4}$$

Add 4 to -32.

$$x=\frac{-2±\sqrt{-28}}{2\times 4}$$

Take the square root of -28.

$$x=\frac{-2±2\sqrt{7}i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-2±2\sqrt{7}i}{8}$$

Now solve the equation x=\frac{-2±2\sqrt{7}i}{8} when ± is plus. Add -2 to 2i\sqrt{7}.

$$x=\frac{-2+2\sqrt{7}i}{8}$$

Divide -2+2i\sqrt{7} by 8.

$$x=\frac{-1+\sqrt{7}i}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{7}i}{8} when ± is minus. Subtract 2i\sqrt{7} from -2.

$$x=\frac{-2\sqrt{7}i-2}{8}$$

Divide -2-2i\sqrt{7} by 8.

$$x=\frac{-\sqrt{7}i-1}{4}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i-1}{4}$$

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}.

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

Calculate 1 to the power of 2 and get 1.

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

Subtract 2x from both sides.

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

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

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

Subtract 1 from both sides.

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

Subtract 1 from -1 to get -2.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{1}{2}+\frac{1}{16}$$

Add -\frac{1}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{7}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{7}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{7}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{7}i}{4} x+\frac{1}{4}=-\frac{\sqrt{7}i}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{-1+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i-1}{4}$$