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

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

Solve for x

$$x=2$$

$$x=0$$

Steps Using Factoring

Subtract \frac{1}{2}x from both sides.

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

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

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

Subtract 2 from both sides.

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

Subtract 2 from 2 to get 0.

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

Factor out x.

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

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

$$x=0 x=2$$

Steps Using the Quadratic Formula

Subtract \frac{1}{2}x from both sides.

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

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

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

Subtract 2 from both sides.

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

Subtract 2 from 2 to get 0.

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

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

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

Take the square root of 1^{2}.

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

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

$$x=\frac{-1±1}{-1}$$

Now solve the equation x=\frac{-1±1}{-1} when ± is plus. Add -1 to 1.

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

Divide 0 by -1.

$$x=0$$

Now solve the equation x=\frac{-1±1}{-1} when ± is minus. Subtract 1 from -1.

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

Divide -2 by -1.

$$x=2$$

The equation is now solved.

$$x=0 x=2$$

Steps for Completing the Square

Subtract \frac{1}{2}x from both sides.

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

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

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

Subtract 2 from both sides.

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

Subtract 2 from 2 to get 0.

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

Multiply both sides by -2.

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

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

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

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

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

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

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

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

$$x^{2}-2x+1=1$$

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

$$\left(x-1\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-1=1 x-1=-1$$

Add 1 to both sides of the equation.

$$x=2 x=0$$