Solve 2x^2-3x=+2x-2x^2

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

Solve for x

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

$$x=0$$

Steps Using Factoring

Subtract 2x from both sides.

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

Combine -3x and -2x to get -5x.

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

Add 2x^{2} to both sides.

$$2x^{2}-5x+2x^{2}=0$$

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

$$4x^{2}-5x=0$$

Factor out x.

$$x\left(4x-5\right)=0$$

To find equation solutions, solve x=0 and 4x-5=0.

$$x=0 x=\frac{5}{4}$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

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

Combine -3x and -2x to get -5x.

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

Add 2x^{2} to both sides.

$$2x^{2}-5x+2x^{2}=0$$

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

$$4x^{2}-5x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2\times 4}$$

Take the square root of \left(-5\right)^{2}.

$$x=\frac{-\left(-5\right)±5}{2\times 4}$$

The opposite of -5 is 5.

$$x=\frac{5±5}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{5±5}{8}$$

Now solve the equation x=\frac{5±5}{8} when ± is plus. Add 5 to 5.

$$x=\frac{10}{8}$$

Reduce the fraction \frac{10}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{4}$$

Now solve the equation x=\frac{5±5}{8} when ± is minus. Subtract 5 from 5.

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

Divide 0 by 8.

$$x=0$$

The equation is now solved.

$$x=\frac{5}{4} x=0$$

Steps for Completing the Square

Subtract 2x from both sides.

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

Combine -3x and -2x to get -5x.

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

Add 2x^{2} to both sides.

$$2x^{2}-5x+2x^{2}=0$$

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

$$4x^{2}-5x=0$$

Divide both sides by 4.

$$\frac{4x^{2}-5x}{4}=\frac{0}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{5}{4}x=\frac{0}{4}$$

Divide 0 by 4.

$$x^{2}-\frac{5}{4}x=0$$

Divide -\frac{5}{4}, the coefficient of the x term, by 2 to get -\frac{5}{8}. Then add the square of -\frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{4}x+\left(-\frac{5}{8}\right)^{2}=\left(-\frac{5}{8}\right)^{2}$$

Square -\frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=\frac{25}{64}$$

Factor x^{2}-\frac{5}{4}x+\frac{25}{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}.

$$\left(x-\frac{5}{8}\right)^{2}=\frac{25}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Simplify.

$$x-\frac{5}{8}=\frac{5}{8} x-\frac{5}{8}=-\frac{5}{8}$$

Add \frac{5}{8} to both sides of the equation.

$$x=\frac{5}{4} x=0$$