Solve 4x^2=2x-4

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

Solve for x (complex solution)

$$x=\frac{1+\sqrt{15}i}{4}\approx 0.25+0.968245837i$$

$$x=\frac{-\sqrt{15}i+1}{4}\approx 0.25-0.968245837i$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

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

Add 4 to both sides.

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

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

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

Square -2.

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

Multiply -4 times 4.

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

Multiply -16 times 4.

$$x=\frac{-\left(-2\right)±\sqrt{4-64}}{2\times 4}$$

Add 4 to -64.

$$x=\frac{-\left(-2\right)±\sqrt{-60}}{2\times 4}$$

Take the square root of -60.

$$x=\frac{-\left(-2\right)±2\sqrt{15}i}{2\times 4}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{15}i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{2±2\sqrt{15}i}{8}$$

Now solve the equation x=\frac{2±2\sqrt{15}i}{8} when ± is plus. Add 2 to 2i\sqrt{15}.

$$x=\frac{2+2\sqrt{15}i}{8}$$

Divide 2+2i\sqrt{15} by 8.

$$x=\frac{1+\sqrt{15}i}{4}$$

Now solve the equation x=\frac{2±2\sqrt{15}i}{8} when ± is minus. Subtract 2i\sqrt{15} from 2.

$$x=\frac{-2\sqrt{15}i+2}{8}$$

Divide 2-2i\sqrt{15} by 8.

$$x=\frac{-\sqrt{15}i+1}{4}$$

The equation is now solved.

$$x=\frac{1+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i+1}{4}$$

Steps for Completing the Square

Subtract 2x from both sides.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

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

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

Divide -4 by 4.

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

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.

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

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

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

Add -1 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{15}{16}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{15}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{15}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{15}i}{4} x-\frac{1}{4}=-\frac{\sqrt{15}i}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{1+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i+1}{4}$$