Solve 4x^2=4x-26

$$4 x ^ { 2 } = 4 x - 26$$

Solve for x (complex solution)

$$x=\frac{1}{2}+\frac{5}{2}i=0.5+2.5i$$

$$x=\frac{1}{2}-\frac{5}{2}i=0.5-2.5i$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$4x^{2}-4x=-26$$

Add 26 to both sides.

$$4x^{2}-4x+26=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -4 for b, and 26 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4\times 26}}{2\times 4}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 4\times 26}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16-16\times 26}}{2\times 4}$$

Multiply -16 times 26.

$$x=\frac{-\left(-4\right)±\sqrt{16-416}}{2\times 4}$$

Add 16 to -416.

$$x=\frac{-\left(-4\right)±\sqrt{-400}}{2\times 4}$$

Take the square root of -400.

$$x=\frac{-\left(-4\right)±20i}{2\times 4}$$

The opposite of -4 is 4.

$$x=\frac{4±20i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{4±20i}{8}$$

Now solve the equation x=\frac{4±20i}{8} when ± is plus. Add 4 to 20i.

$$x=\frac{4+20i}{8}$$

Divide 4+20i by 8.

$$x=\frac{1}{2}+\frac{5}{2}i$$

Now solve the equation x=\frac{4±20i}{8} when ± is minus. Subtract 20i from 4.

$$x=\frac{4-20i}{8}$$

Divide 4-20i by 8.

$$x=\frac{1}{2}-\frac{5}{2}i$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 4x from both sides.

$$4x^{2}-4x=-26$$

Divide both sides by 4.

$$\frac{4x^{2}-4x}{4}=-\frac{26}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{4}{4}\right)x=-\frac{26}{4}$$

Divide -4 by 4.

$$x^{2}-x=-\frac{26}{4}$$

Reduce the fraction \frac{-26}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-x=-\frac{13}{2}$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{13}{2}+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=-\frac{13}{2}+\frac{1}{4}$$

Add -\frac{13}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=-\frac{25}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{25}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{5}{2}i x-\frac{1}{2}=-\frac{5}{2}i$$

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

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