Solve -3/4x+2=-1/2x^2-3/2x+2

$$- \frac { 3 } { 4 } x + 2 = - \frac { 1 } { 2 } x ^ { 2 } - \frac { 3 } { 2 } x + 2$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=0$$

Steps Using Factoring

Add \frac{1}{2}x^{2} to both sides.

$$-\frac{3}{4}x+2+\frac{1}{2}x^{2}=-\frac{3}{2}x+2$$

Add \frac{3}{2}x to both sides.

$$-\frac{3}{4}x+2+\frac{1}{2}x^{2}+\frac{3}{2}x=2$$

Combine -\frac{3}{4}x and \frac{3}{2}x to get \frac{3}{4}x.

$$\frac{3}{4}x+2+\frac{1}{2}x^{2}=2$$

Subtract 2 from both sides.

$$\frac{3}{4}x+2+\frac{1}{2}x^{2}-2=0$$

Subtract 2 from 2 to get 0.

$$\frac{3}{4}x+\frac{1}{2}x^{2}=0$$

Factor out x.

$$x\left(\frac{3}{4}+\frac{1}{2}x\right)=0$$

To find equation solutions, solve x=0 and \frac{3}{4}+\frac{x}{2}=0.

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

Steps Using the Quadratic Formula

Add \frac{1}{2}x^{2} to both sides.

$$-\frac{3}{4}x+2+\frac{1}{2}x^{2}=-\frac{3}{2}x+2$$

Add \frac{3}{2}x to both sides.

$$-\frac{3}{4}x+2+\frac{1}{2}x^{2}+\frac{3}{2}x=2$$

Combine -\frac{3}{4}x and \frac{3}{2}x to get \frac{3}{4}x.

$$\frac{3}{4}x+2+\frac{1}{2}x^{2}=2$$

Subtract 2 from both sides.

$$\frac{3}{4}x+2+\frac{1}{2}x^{2}-2=0$$

Subtract 2 from 2 to get 0.

$$\frac{3}{4}x+\frac{1}{2}x^{2}=0$$

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.

$$\frac{1}{2}x^{2}+\frac{3}{4}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, \frac{3}{4} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{3}{4}±\sqrt{\left(\frac{3}{4}\right)^{2}}}{2\times \frac{1}{2}}$$

Take the square root of \left(\frac{3}{4}\right)^{2}.

$$x=\frac{-\frac{3}{4}±\frac{3}{4}}{2\times \frac{1}{2}}$$

Multiply 2 times \frac{1}{2}.

$$x=\frac{-\frac{3}{4}±\frac{3}{4}}{1}$$

Now solve the equation x=\frac{-\frac{3}{4}±\frac{3}{4}}{1} when ± is plus. Add -\frac{3}{4} to \frac{3}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{1}$$

Divide 0 by 1.

$$x=0$$

Now solve the equation x=\frac{-\frac{3}{4}±\frac{3}{4}}{1} when ± is minus. Subtract \frac{3}{4} from -\frac{3}{4} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{3}{2}}{1}$$

Divide -\frac{3}{2} by 1.

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

The equation is now solved.

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

Steps for Completing the Square

Add \frac{1}{2}x^{2} to both sides.

$$-\frac{3}{4}x+2+\frac{1}{2}x^{2}=-\frac{3}{2}x+2$$

Add \frac{3}{2}x to both sides.

$$-\frac{3}{4}x+2+\frac{1}{2}x^{2}+\frac{3}{2}x=2$$

Combine -\frac{3}{4}x and \frac{3}{2}x to get \frac{3}{4}x.

$$\frac{3}{4}x+2+\frac{1}{2}x^{2}=2$$

Subtract 2 from both sides.

$$\frac{3}{4}x+\frac{1}{2}x^{2}=2-2$$

Subtract 2 from 2 to get 0.

$$\frac{3}{4}x+\frac{1}{2}x^{2}=0$$

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.

$$\frac{1}{2}x^{2}+\frac{3}{4}x=0$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}+\frac{3}{4}x}{\frac{1}{2}}=\frac{0}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$x^{2}+\frac{\frac{3}{4}}{\frac{1}{2}}x=\frac{0}{\frac{1}{2}}$$

Divide \frac{3}{4} by \frac{1}{2} by multiplying \frac{3}{4} by the reciprocal of \frac{1}{2}.

$$x^{2}+\frac{3}{2}x=\frac{0}{\frac{1}{2}}$$

Divide 0 by \frac{1}{2} by multiplying 0 by the reciprocal of \frac{1}{2}.

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

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

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

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{9}{16}$$

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

$$\left(x+\frac{3}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

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

Subtract \frac{3}{4} from both sides of the equation.

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