Solve 4x^2-x=2x

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

Solve for x

$$x=\frac{3}{4}=0.75$$

$$x=0$$

Steps Using Factoring

Subtract 2x from both sides.

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

Combine -x and -2x to get -3x.

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

Factor out x.

$$x\left(4x-3\right)=0$$

To find equation solutions, solve x=0 and 4x-3=0.

$$x=0 x=\frac{3}{4}$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

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

Combine -x and -2x to get -3x.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-3\right)^{2}.

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

The opposite of -3 is 3.

$$x=\frac{3±3}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{3±3}{8}$$

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

$$x=\frac{6}{8}$$

Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{4}$$

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

$$x=\frac{0}{8}$$

Divide 0 by 8.

$$x=0$$

The equation is now solved.

$$x=\frac{3}{4} x=0$$

Steps for Completing the Square

Subtract 2x from both sides.

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

Combine -x and -2x to get -3x.

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

Divide both sides by 4.

$$\frac{4x^{2}-3x}{4}=\frac{0}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{3}{4}x=\frac{0}{4}$$

Divide 0 by 4.

$$x^{2}-\frac{3}{4}x=0$$

Divide -\frac{3}{4}, the coefficient of the x term, by 2 to get -\frac{3}{8}. Then add the square of -\frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{4}x+\left(-\frac{3}{8}\right)^{2}=\left(-\frac{3}{8}\right)^{2}$$

Square -\frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{9}{64}$$

Factor x^{2}-\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{8}\right)^{2}=\frac{9}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{8}\right)^{2}}=\sqrt{\frac{9}{64}}$$

Simplify.

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

Add \frac{3}{8} to both sides of the equation.

$$x=\frac{3}{4} x=0$$