Solve 1/2(3-t)5/3t4/5=9/2-6

$$\frac { 1 } { 2 } ( 3 - t ) \frac { 5 } { 3 } t \frac { 4 } { 5 } = \frac { 9 } { 2 } - 6$$

Solve for t

$$t = \frac{3 \sqrt{2} + 3}{2} \approx 3.621320344$$

$$t=\frac{3-3\sqrt{2}}{2}\approx -0.621320344$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} times \frac{5}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 5}{2\times 3}\left(3-t\right)t\times \frac{4}{5}=\frac{9}{2}-6$$

Do the multiplications in the fraction \frac{1\times 5}{2\times 3}.

$$\frac{5}{6}\left(3-t\right)t\times \frac{4}{5}=\frac{9}{2}-6$$

Multiply \frac{5}{6} times \frac{4}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{5\times 4}{6\times 5}\left(3-t\right)t=\frac{9}{2}-6$$

Cancel out 5 in both numerator and denominator.

$$\frac{4}{6}\left(3-t\right)t=\frac{9}{2}-6$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$\frac{2}{3}\left(3-t\right)t=\frac{9}{2}-6$$

Use the distributive property to multiply \frac{2}{3} by 3-t.

$$\left(\frac{2}{3}\times 3+\frac{2}{3}\left(-1\right)t\right)t=\frac{9}{2}-6$$

Cancel out 3 and 3.

$$\left(2+\frac{2}{3}\left(-1\right)t\right)t=\frac{9}{2}-6$$

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

$$\left(2-\frac{2}{3}t\right)t=\frac{9}{2}-6$$

Use the distributive property to multiply 2-\frac{2}{3}t by t.

$$2t-\frac{2}{3}tt=\frac{9}{2}-6$$

Multiply t and t to get t^{2}.

$$2t-\frac{2}{3}t^{2}=\frac{9}{2}-6$$

Convert 6 to fraction \frac{12}{2}.

$$2t-\frac{2}{3}t^{2}=\frac{9}{2}-\frac{12}{2}$$

Since \frac{9}{2} and \frac{12}{2} have the same denominator, subtract them by subtracting their numerators.

$$2t-\frac{2}{3}t^{2}=\frac{9-12}{2}$$

Subtract 12 from 9 to get -3.

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

Add \frac{3}{2} to both sides.

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

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.

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

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

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

Square 2.

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

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

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

Multiply \frac{8}{3} times \frac{3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

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

Add 4 to 4.

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

Take the square root of 8.

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

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

$$t=\frac{-2±2\sqrt{2}}{-\frac{4}{3}}$$

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

$$t=\frac{2\sqrt{2}-2}{-\frac{4}{3}}$$

Divide -2+2\sqrt{2} by -\frac{4}{3} by multiplying -2+2\sqrt{2} by the reciprocal of -\frac{4}{3}.

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

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

$$t=\frac{-2\sqrt{2}-2}{-\frac{4}{3}}$$

Divide -2-2\sqrt{2} by -\frac{4}{3} by multiplying -2-2\sqrt{2} by the reciprocal of -\frac{4}{3}.

$$t=\frac{3\sqrt{2}+3}{2}$$

The equation is now solved.

$$t=\frac{3-3\sqrt{2}}{2} t=\frac{3\sqrt{2}+3}{2}$$

Steps for Completing the Square

Multiply \frac{1}{2} times \frac{5}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 5}{2\times 3}\left(3-t\right)t\times \frac{4}{5}=\frac{9}{2}-6$$

Do the multiplications in the fraction \frac{1\times 5}{2\times 3}.

$$\frac{5}{6}\left(3-t\right)t\times \frac{4}{5}=\frac{9}{2}-6$$

Multiply \frac{5}{6} times \frac{4}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{5\times 4}{6\times 5}\left(3-t\right)t=\frac{9}{2}-6$$

Cancel out 5 in both numerator and denominator.

$$\frac{4}{6}\left(3-t\right)t=\frac{9}{2}-6$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$\frac{2}{3}\left(3-t\right)t=\frac{9}{2}-6$$

Use the distributive property to multiply \frac{2}{3} by 3-t.

$$\left(\frac{2}{3}\times 3+\frac{2}{3}\left(-1\right)t\right)t=\frac{9}{2}-6$$

Cancel out 3 and 3.

$$\left(2+\frac{2}{3}\left(-1\right)t\right)t=\frac{9}{2}-6$$

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

$$\left(2-\frac{2}{3}t\right)t=\frac{9}{2}-6$$

Use the distributive property to multiply 2-\frac{2}{3}t by t.

$$2t-\frac{2}{3}tt=\frac{9}{2}-6$$

Multiply t and t to get t^{2}.

$$2t-\frac{2}{3}t^{2}=\frac{9}{2}-6$$

Convert 6 to fraction \frac{12}{2}.

$$2t-\frac{2}{3}t^{2}=\frac{9}{2}-\frac{12}{2}$$

Since \frac{9}{2} and \frac{12}{2} have the same denominator, subtract them by subtracting their numerators.

$$2t-\frac{2}{3}t^{2}=\frac{9-12}{2}$$

Subtract 12 from 9 to get -3.

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

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.

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

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

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

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

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

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

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

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

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

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.

$$t^{2}-3t+\left(-\frac{3}{2}\right)^{2}=\frac{9}{4}+\left(-\frac{3}{2}\right)^{2}$$

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

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

Add \frac{9}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

$$\left(t-\frac{3}{2}\right)^{2}=\frac{9}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{2}}$$

Simplify.

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

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

$$t=\frac{3\sqrt{2}+3}{2} t=\frac{3-3\sqrt{2}}{2}$$