Solve 3/2m=-m^2+2m

$$\frac { 3 } { 2 } m = - m ^ { 2 } + 2 m$$

Solve for m

$$m=\frac{1}{2}=0.5$$

$$m=0$$

Steps Using Factoring

Add m^{2} to both sides.

$$\frac{3}{2}m+m^{2}=2m$$

Subtract 2m from both sides.

$$\frac{3}{2}m+m^{2}-2m=0$$

Combine \frac{3}{2}m and -2m to get -\frac{1}{2}m.

$$-\frac{1}{2}m+m^{2}=0$$

Factor out m.

$$m\left(-\frac{1}{2}+m\right)=0$$

To find equation solutions, solve m=0 and -\frac{1}{2}+m=0.

$$m=0 m=\frac{1}{2}$$

Steps Using the Quadratic Formula

Add m^{2} to both sides.

$$\frac{3}{2}m+m^{2}=2m$$

Subtract 2m from both sides.

$$\frac{3}{2}m+m^{2}-2m=0$$

Combine \frac{3}{2}m and -2m to get -\frac{1}{2}m.

$$-\frac{1}{2}m+m^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$m^{2}-\frac{1}{2}m=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{1}{2} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\left(-\frac{1}{2}\right)^{2}}}{2}$$

Take the square root of \left(-\frac{1}{2}\right)^{2}.

$$m=\frac{-\left(-\frac{1}{2}\right)±\frac{1}{2}}{2}$$

The opposite of -\frac{1}{2} is \frac{1}{2}.

$$m=\frac{\frac{1}{2}±\frac{1}{2}}{2}$$

Now solve the equation m=\frac{\frac{1}{2}±\frac{1}{2}}{2} when ± is plus. Add \frac{1}{2} to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$m=\frac{1}{2}$$

Now solve the equation m=\frac{\frac{1}{2}±\frac{1}{2}}{2} when ± is minus. Subtract \frac{1}{2} from \frac{1}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$m=\frac{0}{2}$$

Divide 0 by 2.

$$m=0$$

The equation is now solved.

$$m=\frac{1}{2} m=0$$

Steps for Completing the Square

Add m^{2} to both sides.

$$\frac{3}{2}m+m^{2}=2m$$

Subtract 2m from both sides.

$$\frac{3}{2}m+m^{2}-2m=0$$

Combine \frac{3}{2}m and -2m to get -\frac{1}{2}m.

$$-\frac{1}{2}m+m^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$m^{2}-\frac{1}{2}m=0$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}-\frac{1}{2}m+\left(-\frac{1}{4}\right)^{2}=\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$m^{2}-\frac{1}{2}m+\frac{1}{16}=\frac{1}{16}$$

Factor m^{2}-\frac{1}{2}m+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(m-\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$m-\frac{1}{4}=\frac{1}{4} m-\frac{1}{4}=-\frac{1}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$m=\frac{1}{2} m=0$$