Solve (x)=x^2-2x/2+3

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

Solve for x

$$x=7$$

$$x=0$$

Steps Using Factoring

Add 2 and 3 to get 5.

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

Divide each term of x^{2}-2x by 5 to get \frac{1}{5}x^{2}-\frac{2}{5}x.

$$x=\frac{1}{5}x^{2}-\frac{2}{5}x$$

Subtract \frac{1}{5}x^{2} from both sides.

$$x-\frac{1}{5}x^{2}=-\frac{2}{5}x$$

Add \frac{2}{5}x to both sides.

$$x-\frac{1}{5}x^{2}+\frac{2}{5}x=0$$

Combine x and \frac{2}{5}x to get \frac{7}{5}x.

$$\frac{7}{5}x-\frac{1}{5}x^{2}=0$$

Factor out x.

$$x\left(\frac{7}{5}-\frac{1}{5}x\right)=0$$

To find equation solutions, solve x=0 and \frac{7-x}{5}=0.

$$x=0 x=7$$

Steps Using the Quadratic Formula

Add 2 and 3 to get 5.

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

Divide each term of x^{2}-2x by 5 to get \frac{1}{5}x^{2}-\frac{2}{5}x.

$$x=\frac{1}{5}x^{2}-\frac{2}{5}x$$

Subtract \frac{1}{5}x^{2} from both sides.

$$x-\frac{1}{5}x^{2}=-\frac{2}{5}x$$

Add \frac{2}{5}x to both sides.

$$x-\frac{1}{5}x^{2}+\frac{2}{5}x=0$$

Combine x and \frac{2}{5}x to get \frac{7}{5}x.

$$\frac{7}{5}x-\frac{1}{5}x^{2}=0$$

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.

$$-\frac{1}{5}x^{2}+\frac{7}{5}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{5} for a, \frac{7}{5} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{7}{5}±\sqrt{\left(\frac{7}{5}\right)^{2}}}{2\left(-\frac{1}{5}\right)}$$

Take the square root of \left(\frac{7}{5}\right)^{2}.

$$x=\frac{-\frac{7}{5}±\frac{7}{5}}{2\left(-\frac{1}{5}\right)}$$

Multiply 2 times -\frac{1}{5}.

$$x=\frac{-\frac{7}{5}±\frac{7}{5}}{-\frac{2}{5}}$$

Now solve the equation x=\frac{-\frac{7}{5}±\frac{7}{5}}{-\frac{2}{5}} when ± is plus. Add -\frac{7}{5} to \frac{7}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 0 by -\frac{2}{5} by multiplying 0 by the reciprocal of -\frac{2}{5}.

$$x=0$$

Now solve the equation x=\frac{-\frac{7}{5}±\frac{7}{5}}{-\frac{2}{5}} when ± is minus. Subtract \frac{7}{5} from -\frac{7}{5} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{14}{5}}{-\frac{2}{5}}$$

Divide -\frac{14}{5} by -\frac{2}{5} by multiplying -\frac{14}{5} by the reciprocal of -\frac{2}{5}.

$$x=7$$

The equation is now solved.

$$x=0 x=7$$

Steps for Completing the Square

Add 2 and 3 to get 5.

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

Divide each term of x^{2}-2x by 5 to get \frac{1}{5}x^{2}-\frac{2}{5}x.

$$x=\frac{1}{5}x^{2}-\frac{2}{5}x$$

Subtract \frac{1}{5}x^{2} from both sides.

$$x-\frac{1}{5}x^{2}=-\frac{2}{5}x$$

Add \frac{2}{5}x to both sides.

$$x-\frac{1}{5}x^{2}+\frac{2}{5}x=0$$

Combine x and \frac{2}{5}x to get \frac{7}{5}x.

$$\frac{7}{5}x-\frac{1}{5}x^{2}=0$$

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.

$$-\frac{1}{5}x^{2}+\frac{7}{5}x=0$$

Multiply both sides by -5.

$$\frac{-\frac{1}{5}x^{2}+\frac{7}{5}x}{-\frac{1}{5}}=\frac{0}{-\frac{1}{5}}$$

Dividing by -\frac{1}{5} undoes the multiplication by -\frac{1}{5}.

$$x^{2}+\frac{\frac{7}{5}}{-\frac{1}{5}}x=\frac{0}{-\frac{1}{5}}$$

Divide \frac{7}{5} by -\frac{1}{5} by multiplying \frac{7}{5} by the reciprocal of -\frac{1}{5}.

$$x^{2}-7x=\frac{0}{-\frac{1}{5}}$$

Divide 0 by -\frac{1}{5} by multiplying 0 by the reciprocal of -\frac{1}{5}.

$$x^{2}-7x=0$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=\frac{49}{4}$$

Factor x^{2}-7x+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{7}{2} x-\frac{7}{2}=-\frac{7}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=7 x=0$$