Solve 0=left(3/4t+34-20right)left(-t+52right)

$$0 = \left( \frac{ 3 }{ 4 } t+34-20 \right) \left( -t+52 \right)$$

Solve for t

$$t=52$$

$$t = -\frac{56}{3} = -18\frac{2}{3} \approx -18.666666667$$

Steps Using the Quadratic Formula

Subtract 20 from 34 to get 14.

$$0=\left(\frac{3}{4}t+14\right)\left(-t+52\right)$$

Apply the distributive property by multiplying each term of \frac{3}{4}t+14 by each term of -t+52.

$$0=\frac{3}{4}t\left(-t\right)+\frac{3}{4}t\times 52+14\left(-t\right)+728$$

Express \frac{3}{4}\times 52 as a single fraction.

$$0=\frac{3}{4}t\left(-t\right)+\frac{3\times 52}{4}t+14\left(-t\right)+728$$

Multiply 3 and 52 to get 156.

$$0=\frac{3}{4}t\left(-t\right)+\frac{156}{4}t+14\left(-t\right)+728$$

Divide 156 by 4 to get 39.

$$0=\frac{3}{4}t\left(-t\right)+39t+14\left(-t\right)+728$$

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

$$\frac{3}{4}t\left(-t\right)+39t+14\left(-t\right)+728=0$$

Multiply \frac{3}{4} and -1 to get -\frac{3}{4}.

$$-\frac{3}{4}tt+39t+14\left(-1\right)t+728=0$$

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

$$-\frac{3}{4}t^{2}+39t+14\left(-1\right)t+728=0$$

Multiply 14 and -1 to get -14.

$$-\frac{3}{4}t^{2}+39t-14t+728=0$$

Combine 39t and -14t to get 25t.

$$-\frac{3}{4}t^{2}+25t+728=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{4} for a, 25 for b, and 728 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-25±\sqrt{25^{2}-4\left(-\frac{3}{4}\right)\times 728}}{2\left(-\frac{3}{4}\right)}$$

Square 25.

$$t=\frac{-25±\sqrt{625-4\left(-\frac{3}{4}\right)\times 728}}{2\left(-\frac{3}{4}\right)}$$

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

$$t=\frac{-25±\sqrt{625+3\times 728}}{2\left(-\frac{3}{4}\right)}$$

Multiply 3 times 728.

$$t=\frac{-25±\sqrt{625+2184}}{2\left(-\frac{3}{4}\right)}$$

Add 625 to 2184.

$$t=\frac{-25±\sqrt{2809}}{2\left(-\frac{3}{4}\right)}$$

Take the square root of 2809.

$$t=\frac{-25±53}{2\left(-\frac{3}{4}\right)}$$

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

$$t=\frac{-25±53}{-\frac{3}{2}}$$

Now solve the equation t=\frac{-25±53}{-\frac{3}{2}} when ± is plus. Add -25 to 53.

$$t=\frac{28}{-\frac{3}{2}}$$

Divide 28 by -\frac{3}{2} by multiplying 28 by the reciprocal of -\frac{3}{2}.

$$t=-\frac{56}{3}$$

Now solve the equation t=\frac{-25±53}{-\frac{3}{2}} when ± is minus. Subtract 53 from -25.

$$t=-\frac{78}{-\frac{3}{2}}$$

Divide -78 by -\frac{3}{2} by multiplying -78 by the reciprocal of -\frac{3}{2}.

$$t=52$$

The equation is now solved.

$$t=-\frac{56}{3} t=52$$

Steps for Completing the Square

Subtract 20 from 34 to get 14.

$$0=\left(\frac{3}{4}t+14\right)\left(-t+52\right)$$

Apply the distributive property by multiplying each term of \frac{3}{4}t+14 by each term of -t+52.

$$0=\frac{3}{4}t\left(-t\right)+\frac{3}{4}t\times 52+14\left(-t\right)+728$$

Express \frac{3}{4}\times 52 as a single fraction.

$$0=\frac{3}{4}t\left(-t\right)+\frac{3\times 52}{4}t+14\left(-t\right)+728$$

Multiply 3 and 52 to get 156.

$$0=\frac{3}{4}t\left(-t\right)+\frac{156}{4}t+14\left(-t\right)+728$$

Divide 156 by 4 to get 39.

$$0=\frac{3}{4}t\left(-t\right)+39t+14\left(-t\right)+728$$

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

$$\frac{3}{4}t\left(-t\right)+39t+14\left(-t\right)+728=0$$

Subtract 728 from both sides. Anything subtracted from zero gives its negation.

$$\frac{3}{4}t\left(-t\right)+39t+14\left(-t\right)=-728$$

Multiply \frac{3}{4} and -1 to get -\frac{3}{4}.

$$-\frac{3}{4}tt+39t+14\left(-1\right)t=-728$$

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

$$-\frac{3}{4}t^{2}+39t+14\left(-1\right)t=-728$$

Multiply 14 and -1 to get -14.

$$-\frac{3}{4}t^{2}+39t-14t=-728$$

Combine 39t and -14t to get 25t.

$$-\frac{3}{4}t^{2}+25t=-728$$

Divide both sides of the equation by -\frac{3}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{4}t^{2}+25t}{-\frac{3}{4}}=-\frac{728}{-\frac{3}{4}}$$

Dividing by -\frac{3}{4} undoes the multiplication by -\frac{3}{4}.

$$t^{2}+\frac{25}{-\frac{3}{4}}t=-\frac{728}{-\frac{3}{4}}$$

Divide 25 by -\frac{3}{4} by multiplying 25 by the reciprocal of -\frac{3}{4}.

$$t^{2}-\frac{100}{3}t=-\frac{728}{-\frac{3}{4}}$$

Divide -728 by -\frac{3}{4} by multiplying -728 by the reciprocal of -\frac{3}{4}.

$$t^{2}-\frac{100}{3}t=\frac{2912}{3}$$

Divide -\frac{100}{3}, the coefficient of the x term, by 2 to get -\frac{50}{3}. Then add the square of -\frac{50}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{100}{3}t+\left(-\frac{50}{3}\right)^{2}=\frac{2912}{3}+\left(-\frac{50}{3}\right)^{2}$$

Square -\frac{50}{3} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{100}{3}t+\frac{2500}{9}=\frac{2912}{3}+\frac{2500}{9}$$

Add \frac{2912}{3} to \frac{2500}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{100}{3}t+\frac{2500}{9}=\frac{11236}{9}$$

Factor t^{2}-\frac{100}{3}t+\frac{2500}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{50}{3}\right)^{2}=\frac{11236}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{50}{3}\right)^{2}}=\sqrt{\frac{11236}{9}}$$

Simplify.

$$t-\frac{50}{3}=\frac{106}{3} t-\frac{50}{3}=-\frac{106}{3}$$

Add \frac{50}{3} to both sides of the equation.

$$t=52 t=-\frac{56}{3}$$