Solve x^2-x/4=1/5(x/4-1)

$$x ^ { 2 } - \frac { x } { 4 } = \frac { 1 } { 5 } ( \frac { x } { 4 } - 1 )$$

Solve for x (complex solution)

$$x=\frac{3+\sqrt{71}i}{20}\approx 0.15+0.421307489i$$

$$x=\frac{-\sqrt{71}i+3}{20}\approx 0.15-0.421307489i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 20, the least common multiple of 4,5.

$$20x^{2}-5x=4\left(\frac{x}{4}-1\right)$$

Use the distributive property to multiply 4 by \frac{x}{4}-1.

$$20x^{2}-5x=4\times \frac{x}{4}-4$$

Express 4\times \frac{x}{4} as a single fraction.

$$20x^{2}-5x=\frac{4x}{4}-4$$

Cancel out 4 and 4.

$$20x^{2}-5x=x-4$$

Subtract x from both sides.

$$20x^{2}-5x-x=-4$$

Combine -5x and -x to get -6x.

$$20x^{2}-6x=-4$$

Add 4 to both sides.

$$20x^{2}-6x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, -6 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 20\times 4}}{2\times 20}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 20\times 4}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-6\right)±\sqrt{36-80\times 4}}{2\times 20}$$

Multiply -80 times 4.

$$x=\frac{-\left(-6\right)±\sqrt{36-320}}{2\times 20}$$

Add 36 to -320.

$$x=\frac{-\left(-6\right)±\sqrt{-284}}{2\times 20}$$

Take the square root of -284.

$$x=\frac{-\left(-6\right)±2\sqrt{71}i}{2\times 20}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{71}i}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{6±2\sqrt{71}i}{40}$$

Now solve the equation x=\frac{6±2\sqrt{71}i}{40} when ± is plus. Add 6 to 2i\sqrt{71}.

$$x=\frac{6+2\sqrt{71}i}{40}$$

Divide 6+2i\sqrt{71} by 40.

$$x=\frac{3+\sqrt{71}i}{20}$$

Now solve the equation x=\frac{6±2\sqrt{71}i}{40} when ± is minus. Subtract 2i\sqrt{71} from 6.

$$x=\frac{-2\sqrt{71}i+6}{40}$$

Divide 6-2i\sqrt{71} by 40.

$$x=\frac{-\sqrt{71}i+3}{20}$$

The equation is now solved.

$$x=\frac{3+\sqrt{71}i}{20} x=\frac{-\sqrt{71}i+3}{20}$$

Steps for Completing the Square

Multiply both sides of the equation by 20, the least common multiple of 4,5.

$$20x^{2}-5x=4\left(\frac{x}{4}-1\right)$$

Use the distributive property to multiply 4 by \frac{x}{4}-1.

$$20x^{2}-5x=4\times \frac{x}{4}-4$$

Express 4\times \frac{x}{4} as a single fraction.

$$20x^{2}-5x=\frac{4x}{4}-4$$

Cancel out 4 and 4.

$$20x^{2}-5x=x-4$$

Subtract x from both sides.

$$20x^{2}-5x-x=-4$$

Combine -5x and -x to get -6x.

$$20x^{2}-6x=-4$$

Divide both sides by 20.

$$\frac{20x^{2}-6x}{20}=-\frac{4}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}+\left(-\frac{6}{20}\right)x=-\frac{4}{20}$$

Reduce the fraction \frac{-6}{20} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{3}{10}x=-\frac{4}{20}$$

Reduce the fraction \frac{-4}{20} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{3}{10}x=-\frac{1}{5}$$

Divide -\frac{3}{10}, the coefficient of the x term, by 2 to get -\frac{3}{20}. Then add the square of -\frac{3}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{10}x+\left(-\frac{3}{20}\right)^{2}=-\frac{1}{5}+\left(-\frac{3}{20}\right)^{2}$$

Square -\frac{3}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{10}x+\frac{9}{400}=-\frac{1}{5}+\frac{9}{400}$$

Add -\frac{1}{5} to \frac{9}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{10}x+\frac{9}{400}=-\frac{71}{400}$$

Factor x^{2}-\frac{3}{10}x+\frac{9}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{20}\right)^{2}=-\frac{71}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{20}\right)^{2}}=\sqrt{-\frac{71}{400}}$$

Simplify.

$$x-\frac{3}{20}=\frac{\sqrt{71}i}{20} x-\frac{3}{20}=-\frac{\sqrt{71}i}{20}$$

Add \frac{3}{20} to both sides of the equation.

$$x=\frac{3+\sqrt{71}i}{20} x=\frac{-\sqrt{71}i+3}{20}$$