Solve 18=2/3t^2-2t+6

$$18 = \frac { 2 } { 3 } t ^ { 2 } - 2 t + 6$$

Solve for t

$$t=-3$$

$$t=6$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$\frac{2}{3}t^{2}-2t+6=18$$

Subtract 18 from both sides.

$$\frac{2}{3}t^{2}-2t+6-18=0$$

Subtract 18 from 6 to get -12.

$$\frac{2}{3}t^{2}-2t-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{2}{3} for a, -2 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times \frac{2}{3}\left(-12\right)}}{2\times \frac{2}{3}}$$

Square -2.

$$t=\frac{-\left(-2\right)±\sqrt{4-4\times \frac{2}{3}\left(-12\right)}}{2\times \frac{2}{3}}$$

Multiply -4 times \frac{2}{3}.

$$t=\frac{-\left(-2\right)±\sqrt{4-\frac{8}{3}\left(-12\right)}}{2\times \frac{2}{3}}$$

Multiply -\frac{8}{3} times -12.

$$t=\frac{-\left(-2\right)±\sqrt{4+32}}{2\times \frac{2}{3}}$$

Add 4 to 32.

$$t=\frac{-\left(-2\right)±\sqrt{36}}{2\times \frac{2}{3}}$$

Take the square root of 36.

$$t=\frac{-\left(-2\right)±6}{2\times \frac{2}{3}}$$

The opposite of -2 is 2.

$$t=\frac{2±6}{2\times \frac{2}{3}}$$

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

$$t=\frac{2±6}{\frac{4}{3}}$$

Now solve the equation t=\frac{2±6}{\frac{4}{3}} when ± is plus. Add 2 to 6.

$$t=\frac{8}{\frac{4}{3}}$$

Divide 8 by \frac{4}{3} by multiplying 8 by the reciprocal of \frac{4}{3}.

$$t=6$$

Now solve the equation t=\frac{2±6}{\frac{4}{3}} when ± is minus. Subtract 6 from 2.

$$t=-\frac{4}{\frac{4}{3}}$$

Divide -4 by \frac{4}{3} by multiplying -4 by the reciprocal of \frac{4}{3}.

$$t=-3$$

The equation is now solved.

$$t=6 t=-3$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$\frac{2}{3}t^{2}-2t+6=18$$

Subtract 6 from both sides.

$$\frac{2}{3}t^{2}-2t=18-6$$

Subtract 6 from 18 to get 12.

$$\frac{2}{3}t^{2}-2t=12$$

Divide both sides of the equation by \frac{2}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{2}{3}t^{2}-2t}{\frac{2}{3}}=\frac{12}{\frac{2}{3}}$$

Dividing by \frac{2}{3} undoes the multiplication by \frac{2}{3}.

$$t^{2}+\left(-\frac{2}{\frac{2}{3}}\right)t=\frac{12}{\frac{2}{3}}$$

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

$$t^{2}-3t=\frac{12}{\frac{2}{3}}$$

Divide 12 by \frac{2}{3} by multiplying 12 by the reciprocal of \frac{2}{3}.

$$t^{2}-3t=18$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-3t+\left(-\frac{3}{2}\right)^{2}=18+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-3t+\frac{9}{4}=18+\frac{9}{4}$$

Add 18 to \frac{9}{4}.

$$t^{2}-3t+\frac{9}{4}=\frac{81}{4}$$

Factor t^{2}-3t+\frac{9}{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}.

$$\left(t-\frac{3}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{3}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$t-\frac{3}{2}=\frac{9}{2} t-\frac{3}{2}=-\frac{9}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$t=6 t=-3$$