Solve 1/2(8-t)6/5t=1/3*24

$$\frac { 1 } { 2 } ( 8 - t ) \frac { 6 } { 5 } t = \frac { 1 } { 3 } \cdot 24$$

Solve for t

$$t=\frac{2\sqrt{6}}{3}+4\approx 5.632993162$$

$$t=-\frac{2\sqrt{6}}{3}+4\approx 2.367006838$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} times \frac{6}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 6}{2\times 5}\left(8-t\right)t=\frac{1}{3}\times 24$$

Do the multiplications in the fraction \frac{1\times 6}{2\times 5}.

$$\frac{6}{10}\left(8-t\right)t=\frac{1}{3}\times 24$$

Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

$$\frac{3}{5}\left(8-t\right)t=\frac{1}{3}\times 24$$

Use the distributive property to multiply \frac{3}{5} by 8-t.

$$\left(\frac{3}{5}\times 8+\frac{3}{5}\left(-1\right)t\right)t=\frac{1}{3}\times 24$$

Express \frac{3}{5}\times 8 as a single fraction.

$$\left(\frac{3\times 8}{5}+\frac{3}{5}\left(-1\right)t\right)t=\frac{1}{3}\times 24$$

Multiply 3 and 8 to get 24.

$$\left(\frac{24}{5}+\frac{3}{5}\left(-1\right)t\right)t=\frac{1}{3}\times 24$$

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

$$\left(\frac{24}{5}-\frac{3}{5}t\right)t=\frac{1}{3}\times 24$$

Use the distributive property to multiply \frac{24}{5}-\frac{3}{5}t by t.

$$\frac{24}{5}t-\frac{3}{5}tt=\frac{1}{3}\times 24$$

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

$$\frac{24}{5}t-\frac{3}{5}t^{2}=\frac{1}{3}\times 24$$

Multiply \frac{1}{3} and 24 to get \frac{24}{3}.

$$\frac{24}{5}t-\frac{3}{5}t^{2}=\frac{24}{3}$$

Divide 24 by 3 to get 8.

$$\frac{24}{5}t-\frac{3}{5}t^{2}=8$$

Subtract 8 from both sides.

$$\frac{24}{5}t-\frac{3}{5}t^{2}-8=0$$

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.

$$-\frac{3}{5}t^{2}+\frac{24}{5}t-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{5} for a, \frac{24}{5} for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\frac{24}{5}±\sqrt{\left(\frac{24}{5}\right)^{2}-4\left(-\frac{3}{5}\right)\left(-8\right)}}{2\left(-\frac{3}{5}\right)}$$

Square \frac{24}{5} by squaring both the numerator and the denominator of the fraction.

$$t=\frac{-\frac{24}{5}±\sqrt{\frac{576}{25}-4\left(-\frac{3}{5}\right)\left(-8\right)}}{2\left(-\frac{3}{5}\right)}$$

Multiply -4 times -\frac{3}{5}.

$$t=\frac{-\frac{24}{5}±\sqrt{\frac{576}{25}+\frac{12}{5}\left(-8\right)}}{2\left(-\frac{3}{5}\right)}$$

Multiply \frac{12}{5} times -8.

$$t=\frac{-\frac{24}{5}±\sqrt{\frac{576}{25}-\frac{96}{5}}}{2\left(-\frac{3}{5}\right)}$$

Add \frac{576}{25} to -\frac{96}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t=\frac{-\frac{24}{5}±\sqrt{\frac{96}{25}}}{2\left(-\frac{3}{5}\right)}$$

Take the square root of \frac{96}{25}.

$$t=\frac{-\frac{24}{5}±\frac{4\sqrt{6}}{5}}{2\left(-\frac{3}{5}\right)}$$

Multiply 2 times -\frac{3}{5}.

$$t=\frac{-\frac{24}{5}±\frac{4\sqrt{6}}{5}}{-\frac{6}{5}}$$

Now solve the equation t=\frac{-\frac{24}{5}±\frac{4\sqrt{6}}{5}}{-\frac{6}{5}} when ± is plus. Add -\frac{24}{5} to \frac{4\sqrt{6}}{5}.

$$t=\frac{4\sqrt{6}-24}{-\frac{6}{5}\times 5}$$

Divide \frac{-24+4\sqrt{6}}{5} by -\frac{6}{5} by multiplying \frac{-24+4\sqrt{6}}{5} by the reciprocal of -\frac{6}{5}.

$$t=-\frac{2\sqrt{6}}{3}+4$$

Now solve the equation t=\frac{-\frac{24}{5}±\frac{4\sqrt{6}}{5}}{-\frac{6}{5}} when ± is minus. Subtract \frac{4\sqrt{6}}{5} from -\frac{24}{5}.

$$t=\frac{-4\sqrt{6}-24}{-\frac{6}{5}\times 5}$$

Divide \frac{-24-4\sqrt{6}}{5} by -\frac{6}{5} by multiplying \frac{-24-4\sqrt{6}}{5} by the reciprocal of -\frac{6}{5}.

$$t=\frac{2\sqrt{6}}{3}+4$$

The equation is now solved.

$$t=-\frac{2\sqrt{6}}{3}+4 t=\frac{2\sqrt{6}}{3}+4$$

Steps for Completing the Square

Multiply \frac{1}{2} times \frac{6}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 6}{2\times 5}\left(8-t\right)t=\frac{1}{3}\times 24$$

Do the multiplications in the fraction \frac{1\times 6}{2\times 5}.

$$\frac{6}{10}\left(8-t\right)t=\frac{1}{3}\times 24$$

Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

$$\frac{3}{5}\left(8-t\right)t=\frac{1}{3}\times 24$$

Use the distributive property to multiply \frac{3}{5} by 8-t.

$$\left(\frac{3}{5}\times 8+\frac{3}{5}\left(-1\right)t\right)t=\frac{1}{3}\times 24$$

Express \frac{3}{5}\times 8 as a single fraction.

$$\left(\frac{3\times 8}{5}+\frac{3}{5}\left(-1\right)t\right)t=\frac{1}{3}\times 24$$

Multiply 3 and 8 to get 24.

$$\left(\frac{24}{5}+\frac{3}{5}\left(-1\right)t\right)t=\frac{1}{3}\times 24$$

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

$$\left(\frac{24}{5}-\frac{3}{5}t\right)t=\frac{1}{3}\times 24$$

Use the distributive property to multiply \frac{24}{5}-\frac{3}{5}t by t.

$$\frac{24}{5}t-\frac{3}{5}tt=\frac{1}{3}\times 24$$

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

$$\frac{24}{5}t-\frac{3}{5}t^{2}=\frac{1}{3}\times 24$$

Multiply \frac{1}{3} and 24 to get \frac{24}{3}.

$$\frac{24}{5}t-\frac{3}{5}t^{2}=\frac{24}{3}$$

Divide 24 by 3 to get 8.

$$\frac{24}{5}t-\frac{3}{5}t^{2}=8$$

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.

$$-\frac{3}{5}t^{2}+\frac{24}{5}t=8$$

Divide both sides of the equation by -\frac{3}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{5}t^{2}+\frac{24}{5}t}{-\frac{3}{5}}=\frac{8}{-\frac{3}{5}}$$

Dividing by -\frac{3}{5} undoes the multiplication by -\frac{3}{5}.

$$t^{2}+\frac{\frac{24}{5}}{-\frac{3}{5}}t=\frac{8}{-\frac{3}{5}}$$

Divide \frac{24}{5} by -\frac{3}{5} by multiplying \frac{24}{5} by the reciprocal of -\frac{3}{5}.

$$t^{2}-8t=\frac{8}{-\frac{3}{5}}$$

Divide 8 by -\frac{3}{5} by multiplying 8 by the reciprocal of -\frac{3}{5}.

$$t^{2}-8t=-\frac{40}{3}$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-8t+\left(-4\right)^{2}=-\frac{40}{3}+\left(-4\right)^{2}$$

Square -4.

$$t^{2}-8t+16=-\frac{40}{3}+16$$

Add -\frac{40}{3} to 16.

$$t^{2}-8t+16=\frac{8}{3}$$

Factor t^{2}-8t+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-4\right)^{2}=\frac{8}{3}$$

Take the square root of both sides of the equation.

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

Simplify.

$$t-4=\frac{2\sqrt{6}}{3} t-4=-\frac{2\sqrt{6}}{3}$$

Add 4 to both sides of the equation.

$$t=\frac{2\sqrt{6}}{3}+4 t=-\frac{2\sqrt{6}}{3}+4$$