Solve 3x+2=-x^2+7/2x+2

$$3 x + 2 = - x ^ { 2 } + \frac { 7 } { 2 } x + 2$$

Solve for x

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

$$x=0$$

Steps Using Factoring

Add x^{2} to both sides.

$$3x+2+x^{2}=\frac{7}{2}x+2$$

Subtract \frac{7}{2}x from both sides.

$$3x+2+x^{2}-\frac{7}{2}x=2$$

Combine 3x and -\frac{7}{2}x to get -\frac{1}{2}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.

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

Factor out x.

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

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

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

Steps Using the Quadratic Formula

Add x^{2} to both sides.

$$3x+2+x^{2}=\frac{7}{2}x+2$$

Subtract \frac{7}{2}x from both sides.

$$3x+2+x^{2}-\frac{7}{2}x=2$$

Combine 3x and -\frac{7}{2}x to get -\frac{1}{2}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.

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

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.

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

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}.

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

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

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

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

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

Now solve the equation x=\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.

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

Now solve the equation x=\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.

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

Divide 0 by 2.

$$x=0$$

The equation is now solved.

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

Steps for Completing the Square

Add x^{2} to both sides.

$$3x+2+x^{2}=\frac{7}{2}x+2$$

Subtract \frac{7}{2}x from both sides.

$$3x+2+x^{2}-\frac{7}{2}x=2$$

Combine 3x and -\frac{7}{2}x to get -\frac{1}{2}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.

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

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.

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

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.

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

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

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

Factor x^{2}-\frac{1}{2}x+\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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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