Solve -1/6(t+2)^2+5/6(t+2)+4=4

$$- \frac { 1 } { 6 } ( t + 2 ) ^ { 2 } + \frac { 5 } { 6 } ( t + 2 ) + 4 = 4$$

Solve for t

$$t=3$$

$$t=-2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+2\right)^{2}.

$$-\frac{1}{6}\left(t^{2}+4t+4\right)+\frac{5}{6}\left(t+2\right)+4=4$$

Use the distributive property to multiply -\frac{1}{6} by t^{2}+4t+4.

$$-\frac{1}{6}t^{2}-\frac{2}{3}t-\frac{2}{3}+\frac{5}{6}\left(t+2\right)+4=4$$

Use the distributive property to multiply \frac{5}{6} by t+2.

$$-\frac{1}{6}t^{2}-\frac{2}{3}t-\frac{2}{3}+\frac{5}{6}t+\frac{5}{3}+4=4$$

Combine -\frac{2}{3}t and \frac{5}{6}t to get \frac{1}{6}t.

$$-\frac{1}{6}t^{2}+\frac{1}{6}t-\frac{2}{3}+\frac{5}{3}+4=4$$

Add -\frac{2}{3} and \frac{5}{3} to get 1.

$$-\frac{1}{6}t^{2}+\frac{1}{6}t+1+4=4$$

Add 1 and 4 to get 5.

$$-\frac{1}{6}t^{2}+\frac{1}{6}t+5=4$$

Subtract 4 from both sides.

$$-\frac{1}{6}t^{2}+\frac{1}{6}t+5-4=0$$

Subtract 4 from 5 to get 1.

$$-\frac{1}{6}t^{2}+\frac{1}{6}t+1=0$$

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

$$t=\frac{-\frac{1}{6}±\sqrt{\left(\frac{1}{6}\right)^{2}-4\left(-\frac{1}{6}\right)}}{2\left(-\frac{1}{6}\right)}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$t=\frac{-\frac{1}{6}±\sqrt{\frac{1}{36}-4\left(-\frac{1}{6}\right)}}{2\left(-\frac{1}{6}\right)}$$

Multiply -4 times -\frac{1}{6}.

$$t=\frac{-\frac{1}{6}±\sqrt{\frac{1}{36}+\frac{2}{3}}}{2\left(-\frac{1}{6}\right)}$$

Add \frac{1}{36} to \frac{2}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t=\frac{-\frac{1}{6}±\sqrt{\frac{25}{36}}}{2\left(-\frac{1}{6}\right)}$$

Take the square root of \frac{25}{36}.

$$t=\frac{-\frac{1}{6}±\frac{5}{6}}{2\left(-\frac{1}{6}\right)}$$

Multiply 2 times -\frac{1}{6}.

$$t=\frac{-\frac{1}{6}±\frac{5}{6}}{-\frac{1}{3}}$$

Now solve the equation t=\frac{-\frac{1}{6}±\frac{5}{6}}{-\frac{1}{3}} when ± is plus. Add -\frac{1}{6} to \frac{5}{6} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t=\frac{\frac{2}{3}}{-\frac{1}{3}}$$

Divide \frac{2}{3} by -\frac{1}{3} by multiplying \frac{2}{3} by the reciprocal of -\frac{1}{3}.

$$t=-2$$

Now solve the equation t=\frac{-\frac{1}{6}±\frac{5}{6}}{-\frac{1}{3}} when ± is minus. Subtract \frac{5}{6} from -\frac{1}{6} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$t=-\frac{1}{-\frac{1}{3}}$$

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

$$t=3$$

The equation is now solved.

$$t=-2 t=3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+2\right)^{2}.

$$-\frac{1}{6}\left(t^{2}+4t+4\right)+\frac{5}{6}\left(t+2\right)+4=4$$

Use the distributive property to multiply -\frac{1}{6} by t^{2}+4t+4.

$$-\frac{1}{6}t^{2}-\frac{2}{3}t-\frac{2}{3}+\frac{5}{6}\left(t+2\right)+4=4$$

Use the distributive property to multiply \frac{5}{6} by t+2.

$$-\frac{1}{6}t^{2}-\frac{2}{3}t-\frac{2}{3}+\frac{5}{6}t+\frac{5}{3}+4=4$$

Combine -\frac{2}{3}t and \frac{5}{6}t to get \frac{1}{6}t.

$$-\frac{1}{6}t^{2}+\frac{1}{6}t-\frac{2}{3}+\frac{5}{3}+4=4$$

Add -\frac{2}{3} and \frac{5}{3} to get 1.

$$-\frac{1}{6}t^{2}+\frac{1}{6}t+1+4=4$$

Add 1 and 4 to get 5.

$$-\frac{1}{6}t^{2}+\frac{1}{6}t+5=4$$

Subtract 5 from both sides.

$$-\frac{1}{6}t^{2}+\frac{1}{6}t=4-5$$

Subtract 5 from 4 to get -1.

$$-\frac{1}{6}t^{2}+\frac{1}{6}t=-1$$

Multiply both sides by -6.

$$\frac{-\frac{1}{6}t^{2}+\frac{1}{6}t}{-\frac{1}{6}}=-\frac{1}{-\frac{1}{6}}$$

Dividing by -\frac{1}{6} undoes the multiplication by -\frac{1}{6}.

$$t^{2}+\frac{\frac{1}{6}}{-\frac{1}{6}}t=-\frac{1}{-\frac{1}{6}}$$

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

$$t^{2}-t=-\frac{1}{-\frac{1}{6}}$$

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

$$t^{2}-t=6$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-t+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$t^{2}-t+\frac{1}{4}=\frac{25}{4}$$

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

$$\left(t-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$t-\frac{1}{2}=\frac{5}{2} t-\frac{1}{2}=-\frac{5}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$t=3 t=-2$$