Solve (2)(2x-1)^2+12x-1=0

$$( 2 ) ( 2 x - 1 ) ^ { 2 } + 12 x - 1 = 0$$

Solve for x (complex solution)

$$x=-\frac{1}{4}+\frac{1}{4}i=-0.25+0.25i$$

$$x=-\frac{1}{4}-\frac{1}{4}i=-0.25-0.25i$$

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}.

$$2\left(4x^{2}-4x+1\right)+12x-1=0$$

Use the distributive property to multiply 2 by 4x^{2}-4x+1.

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

Combine -8x and 12x to get 4x.

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

Subtract 1 from 2 to get 1.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 4 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 4.

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

Multiply -4 times 8.

$$x=\frac{-4±\sqrt{16-32}}{2\times 8}$$

Add 16 to -32.

$$x=\frac{-4±\sqrt{-16}}{2\times 8}$$

Take the square root of -16.

$$x=\frac{-4±4i}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-4±4i}{16}$$

Now solve the equation x=\frac{-4±4i}{16} when ± is plus. Add -4 to 4i.

$$x=\frac{-4+4i}{16}$$

Divide -4+4i by 16.

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

Now solve the equation x=\frac{-4±4i}{16} when ± is minus. Subtract 4i from -4.

$$x=\frac{-4-4i}{16}$$

Divide -4-4i by 16.

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

The equation is now solved.

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

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}.

$$2\left(4x^{2}-4x+1\right)+12x-1=0$$

Use the distributive property to multiply 2 by 4x^{2}-4x+1.

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

Combine -8x and 12x to get 4x.

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

Subtract 1 from 2 to get 1.

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$8x^{2}+4x=-1$$

Divide both sides by 8.

$$\frac{8x^{2}+4x}{8}=-\frac{1}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{4}{8}x=-\frac{1}{8}$$

Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{1}{2}x=-\frac{1}{8}$$

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.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=-\frac{1}{8}+\left(\frac{1}{4}\right)^{2}$$

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

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{1}{8}+\frac{1}{16}$$

Add -\frac{1}{8} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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