Solve -1/3m^2+8/3m+3+1/3(m-1)-3=3

$$- \frac { 1 } { 3 } m ^ { 2 } + \frac { 8 } { 3 } m + 3 + \frac { 1 } { 3 } ( m - 1 ) - 3 = 3$$

Solve for m

$$m = \frac{\sqrt{41} + 9}{2} \approx 7.701562119$$

$$m = \frac{9 - \sqrt{41}}{2} \approx 1.298437881$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{3} by m-1.

$$-\frac{1}{3}m^{2}+\frac{8}{3}m+3+\frac{1}{3}m-\frac{1}{3}-3=3$$

Combine \frac{8}{3}m and \frac{1}{3}m to get 3m.

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

Subtract \frac{1}{3} from 3 to get \frac{8}{3}.

$$-\frac{1}{3}m^{2}+3m+\frac{8}{3}-3=3$$

Subtract 3 from \frac{8}{3} to get -\frac{1}{3}.

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

Subtract 3 from both sides.

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

Subtract 3 from -\frac{1}{3} to get -\frac{10}{3}.

$$-\frac{1}{3}m^{2}+3m-\frac{10}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{3} for a, 3 for b, and -\frac{10}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-3±\sqrt{3^{2}-4\left(-\frac{1}{3}\right)\left(-\frac{10}{3}\right)}}{2\left(-\frac{1}{3}\right)}$$

Square 3.

$$m=\frac{-3±\sqrt{9-4\left(-\frac{1}{3}\right)\left(-\frac{10}{3}\right)}}{2\left(-\frac{1}{3}\right)}$$

Multiply -4 times -\frac{1}{3}.

$$m=\frac{-3±\sqrt{9+\frac{4}{3}\left(-\frac{10}{3}\right)}}{2\left(-\frac{1}{3}\right)}$$

Multiply \frac{4}{3} times -\frac{10}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$m=\frac{-3±\sqrt{9-\frac{40}{9}}}{2\left(-\frac{1}{3}\right)}$$

Add 9 to -\frac{40}{9}.

$$m=\frac{-3±\sqrt{\frac{41}{9}}}{2\left(-\frac{1}{3}\right)}$$

Take the square root of \frac{41}{9}.

$$m=\frac{-3±\frac{\sqrt{41}}{3}}{2\left(-\frac{1}{3}\right)}$$

Multiply 2 times -\frac{1}{3}.

$$m=\frac{-3±\frac{\sqrt{41}}{3}}{-\frac{2}{3}}$$

Now solve the equation m=\frac{-3±\frac{\sqrt{41}}{3}}{-\frac{2}{3}} when ± is plus. Add -3 to \frac{\sqrt{41}}{3}.

$$m=\frac{\frac{\sqrt{41}}{3}-3}{-\frac{2}{3}}$$

Divide -3+\frac{\sqrt{41}}{3} by -\frac{2}{3} by multiplying -3+\frac{\sqrt{41}}{3} by the reciprocal of -\frac{2}{3}.

$$m=\frac{9-\sqrt{41}}{2}$$

Now solve the equation m=\frac{-3±\frac{\sqrt{41}}{3}}{-\frac{2}{3}} when ± is minus. Subtract \frac{\sqrt{41}}{3} from -3.

$$m=\frac{-\frac{\sqrt{41}}{3}-3}{-\frac{2}{3}}$$

Divide -3-\frac{\sqrt{41}}{3} by -\frac{2}{3} by multiplying -3-\frac{\sqrt{41}}{3} by the reciprocal of -\frac{2}{3}.

$$m=\frac{\sqrt{41}+9}{2}$$

The equation is now solved.

$$m=\frac{9-\sqrt{41}}{2} m=\frac{\sqrt{41}+9}{2}$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{3} by m-1.

$$-\frac{1}{3}m^{2}+\frac{8}{3}m+3+\frac{1}{3}m-\frac{1}{3}-3=3$$

Combine \frac{8}{3}m and \frac{1}{3}m to get 3m.

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

Subtract \frac{1}{3} from 3 to get \frac{8}{3}.

$$-\frac{1}{3}m^{2}+3m+\frac{8}{3}-3=3$$

Subtract 3 from \frac{8}{3} to get -\frac{1}{3}.

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

Add \frac{1}{3} to both sides.

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

Add 3 and \frac{1}{3} to get \frac{10}{3}.

$$-\frac{1}{3}m^{2}+3m=\frac{10}{3}$$

Multiply both sides by -3.

$$\frac{-\frac{1}{3}m^{2}+3m}{-\frac{1}{3}}=\frac{\frac{10}{3}}{-\frac{1}{3}}$$

Dividing by -\frac{1}{3} undoes the multiplication by -\frac{1}{3}.

$$m^{2}+\frac{3}{-\frac{1}{3}}m=\frac{\frac{10}{3}}{-\frac{1}{3}}$$

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

$$m^{2}-9m=\frac{\frac{10}{3}}{-\frac{1}{3}}$$

Divide \frac{10}{3} by -\frac{1}{3} by multiplying \frac{10}{3} by the reciprocal of -\frac{1}{3}.

$$m^{2}-9m=-10$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}-9m+\left(-\frac{9}{2}\right)^{2}=-10+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$m^{2}-9m+\frac{81}{4}=-10+\frac{81}{4}$$

Add -10 to \frac{81}{4}.

$$m^{2}-9m+\frac{81}{4}=\frac{41}{4}$$

Factor m^{2}-9m+\frac{81}{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}.

$$\left(m-\frac{9}{2}\right)^{2}=\frac{41}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-\frac{9}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Simplify.

$$m-\frac{9}{2}=\frac{\sqrt{41}}{2} m-\frac{9}{2}=-\frac{\sqrt{41}}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$m=\frac{\sqrt{41}+9}{2} m=\frac{9-\sqrt{41}}{2}$$