Solve m^2-6/5=m

$$\frac { m ^ { 2 } - 6 } { 5 } = m$$

Solve for m

$$m=-1$$

$$m=6$$

Steps Using the Quadratic Formula

Divide each term of m^{2}-6 by 5 to get \frac{1}{5}m^{2}-\frac{6}{5}.

$$\frac{1}{5}m^{2}-\frac{6}{5}=m$$

Subtract m from both sides.

$$\frac{1}{5}m^{2}-\frac{6}{5}-m=0$$

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.

$$\frac{1}{5}m^{2}-m-\frac{6}{5}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{5} for a, -1 for b, and -\frac{6}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{1}{5}\left(-\frac{6}{5}\right)}}{2\times \frac{1}{5}}$$

Multiply -4 times \frac{1}{5}.

$$m=\frac{-\left(-1\right)±\sqrt{1-\frac{4}{5}\left(-\frac{6}{5}\right)}}{2\times \frac{1}{5}}$$

Multiply -\frac{4}{5} times -\frac{6}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$m=\frac{-\left(-1\right)±\sqrt{1+\frac{24}{25}}}{2\times \frac{1}{5}}$$

Add 1 to \frac{24}{25}.

$$m=\frac{-\left(-1\right)±\sqrt{\frac{49}{25}}}{2\times \frac{1}{5}}$$

Take the square root of \frac{49}{25}.

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

The opposite of -1 is 1.

$$m=\frac{1±\frac{7}{5}}{2\times \frac{1}{5}}$$

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

$$m=\frac{1±\frac{7}{5}}{\frac{2}{5}}$$

Now solve the equation m=\frac{1±\frac{7}{5}}{\frac{2}{5}} when ± is plus. Add 1 to \frac{7}{5}.

$$m=\frac{\frac{12}{5}}{\frac{2}{5}}$$

Divide \frac{12}{5} by \frac{2}{5} by multiplying \frac{12}{5} by the reciprocal of \frac{2}{5}.

$$m=6$$

Now solve the equation m=\frac{1±\frac{7}{5}}{\frac{2}{5}} when ± is minus. Subtract \frac{7}{5} from 1.

$$m=-\frac{\frac{2}{5}}{\frac{2}{5}}$$

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

$$m=-1$$

The equation is now solved.

$$m=6 m=-1$$

Steps for Completing the Square

Divide each term of m^{2}-6 by 5 to get \frac{1}{5}m^{2}-\frac{6}{5}.

$$\frac{1}{5}m^{2}-\frac{6}{5}=m$$

Subtract m from both sides.

$$\frac{1}{5}m^{2}-\frac{6}{5}-m=0$$

Add \frac{6}{5} to both sides. Anything plus zero gives itself.

$$\frac{1}{5}m^{2}-m=\frac{6}{5}$$

Multiply both sides by 5.

$$\frac{\frac{1}{5}m^{2}-m}{\frac{1}{5}}=\frac{\frac{6}{5}}{\frac{1}{5}}$$

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

$$m^{2}+\left(-\frac{1}{\frac{1}{5}}\right)m=\frac{\frac{6}{5}}{\frac{1}{5}}$$

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

$$m^{2}-5m=\frac{\frac{6}{5}}{\frac{1}{5}}$$

Divide \frac{6}{5} by \frac{1}{5} by multiplying \frac{6}{5} by the reciprocal of \frac{1}{5}.

$$m^{2}-5m=6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}-5m+\left(-\frac{5}{2}\right)^{2}=6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$m^{2}-5m+\frac{25}{4}=6+\frac{25}{4}$$

Add 6 to \frac{25}{4}.

$$m^{2}-5m+\frac{25}{4}=\frac{49}{4}$$

Factor m^{2}-5m+\frac{25}{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}.

$$\left(m-\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$m-\frac{5}{2}=\frac{7}{2} m-\frac{5}{2}=-\frac{7}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$m=6 m=-1$$