Solve 4x^2-6x=2x^2

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

Solve for x

$$x=3$$

$$x=0$$

Steps Using Factoring

Subtract 2x^{2} from both sides.

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

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

$$2x^{2}-6x=0$$

Factor out x.

$$x\left(2x-6\right)=0$$

To find equation solutions, solve x=0 and 2x-6=0.

$$x=0 x=3$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

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

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

$$2x^{2}-6x=0$$

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

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

Take the square root of \left(-6\right)^{2}.

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

The opposite of -6 is 6.

$$x=\frac{6±6}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{6±6}{4}$$

Now solve the equation x=\frac{6±6}{4} when ± is plus. Add 6 to 6.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{6±6}{4} when ± is minus. Subtract 6 from 6.

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

Divide 0 by 4.

$$x=0$$

The equation is now solved.

$$x=3 x=0$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

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

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

$$2x^{2}-6x=0$$

Divide both sides by 2.

$$\frac{2x^{2}-6x}{2}=\frac{0}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{6}{2}\right)x=\frac{0}{2}$$

Divide -6 by 2.

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

Divide 0 by 2.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=3 x=0$$