Solve -3i^2=(2x-1)^2

$$- 3 i ^ { 2 } = ( 2 x - 1 ) ^ { 2 }$$

Solve for x

$$x=\frac{1-\sqrt{3}}{2}\approx -0.366025404$$

$$x = \frac{\sqrt{3} + 1}{2} \approx 1.366025404$$

Steps Using the Quadratic Formula

Calculate i to the power of 2 and get -1.

$$-3\left(-1\right)=\left(2x-1\right)^{2}$$

Multiply -3 and -1 to get 3.

$$3=\left(2x-1\right)^{2}$$

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

$$3=4x^{2}-4x+1$$

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}-4x+1=3$$

Subtract 3 from both sides.

$$4x^{2}-4x+1-3=0$$

Subtract 3 from 1 to get -2.

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

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

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

Square -4.

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

Multiply -4 times 4.

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

Multiply -16 times -2.

$$x=\frac{-\left(-4\right)±\sqrt{16+32}}{2\times 4}$$

Add 16 to 32.

$$x=\frac{-\left(-4\right)±\sqrt{48}}{2\times 4}$$

Take the square root of 48.

$$x=\frac{-\left(-4\right)±4\sqrt{3}}{2\times 4}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{3}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{4±4\sqrt{3}}{8}$$

Now solve the equation x=\frac{4±4\sqrt{3}}{8} when ± is plus. Add 4 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}+4}{8}$$

Divide 4+4\sqrt{3} by 8.

$$x=\frac{\sqrt{3}+1}{2}$$

Now solve the equation x=\frac{4±4\sqrt{3}}{8} when ± is minus. Subtract 4\sqrt{3} from 4.

$$x=\frac{4-4\sqrt{3}}{8}$$

Divide 4-4\sqrt{3} by 8.

$$x=\frac{1-\sqrt{3}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{3}+1}{2} x=\frac{1-\sqrt{3}}{2}$$

Steps for Completing the Square

Calculate i to the power of 2 and get -1.

$$-3\left(-1\right)=\left(2x-1\right)^{2}$$

Multiply -3 and -1 to get 3.

$$3=\left(2x-1\right)^{2}$$

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

$$3=4x^{2}-4x+1$$

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}-4x+1=3$$

Subtract 1 from both sides.

$$4x^{2}-4x=3-1$$

Subtract 1 from 3 to get 2.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide -4 by 4.

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

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

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{1}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{3}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{3}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{3}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{3}}{2} x-\frac{1}{2}=-\frac{\sqrt{3}}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{3}+1}{2} x=\frac{1-\sqrt{3}}{2}$$