Solve 75+10t=30t-1/2*2.5t^2

$$75 + 10 t = 30 t - \frac { 1 } { 2 } \times 2.5 t ^ { 2 }$$

Solve for t

$$t=6$$

$$t=10$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} and 2.5 to get \frac{5}{4}.

$$75+10t=30t-\frac{5}{4}t^{2}$$

Subtract 30t from both sides.

$$75+10t-30t=-\frac{5}{4}t^{2}$$

Combine 10t and -30t to get -20t.

$$75-20t=-\frac{5}{4}t^{2}$$

Add \frac{5}{4}t^{2} to both sides.

$$75-20t+\frac{5}{4}t^{2}=0$$

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.

$$\frac{5}{4}t^{2}-20t+75=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{5}{4} for a, -20 for b, and 75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times \frac{5}{4}\times 75}}{2\times \frac{5}{4}}$$

Square -20.

$$t=\frac{-\left(-20\right)±\sqrt{400-4\times \frac{5}{4}\times 75}}{2\times \frac{5}{4}}$$

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

$$t=\frac{-\left(-20\right)±\sqrt{400-5\times 75}}{2\times \frac{5}{4}}$$

Multiply -5 times 75.

$$t=\frac{-\left(-20\right)±\sqrt{400-375}}{2\times \frac{5}{4}}$$

Add 400 to -375.

$$t=\frac{-\left(-20\right)±\sqrt{25}}{2\times \frac{5}{4}}$$

Take the square root of 25.

$$t=\frac{-\left(-20\right)±5}{2\times \frac{5}{4}}$$

The opposite of -20 is 20.

$$t=\frac{20±5}{2\times \frac{5}{4}}$$

Multiply 2 times \frac{5}{4}.

$$t=\frac{20±5}{\frac{5}{2}}$$

Now solve the equation t=\frac{20±5}{\frac{5}{2}} when ± is plus. Add 20 to 5.

$$t=\frac{25}{\frac{5}{2}}$$

Divide 25 by \frac{5}{2} by multiplying 25 by the reciprocal of \frac{5}{2}.

$$t=10$$

Now solve the equation t=\frac{20±5}{\frac{5}{2}} when ± is minus. Subtract 5 from 20.

$$t=\frac{15}{\frac{5}{2}}$$

Divide 15 by \frac{5}{2} by multiplying 15 by the reciprocal of \frac{5}{2}.

$$t=6$$

The equation is now solved.

$$t=10 t=6$$

Steps for Completing the Square

Multiply \frac{1}{2} and 2.5 to get \frac{5}{4}.

$$75+10t=30t-\frac{5}{4}t^{2}$$

Subtract 30t from both sides.

$$75+10t-30t=-\frac{5}{4}t^{2}$$

Combine 10t and -30t to get -20t.

$$75-20t=-\frac{5}{4}t^{2}$$

Add \frac{5}{4}t^{2} to both sides.

$$75-20t+\frac{5}{4}t^{2}=0$$

Subtract 75 from both sides. Anything subtracted from zero gives its negation.

$$-20t+\frac{5}{4}t^{2}=-75$$

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.

$$\frac{5}{4}t^{2}-20t=-75$$

Divide both sides of the equation by \frac{5}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{4}t^{2}-20t}{\frac{5}{4}}=-\frac{75}{\frac{5}{4}}$$

Dividing by \frac{5}{4} undoes the multiplication by \frac{5}{4}.

$$t^{2}+\left(-\frac{20}{\frac{5}{4}}\right)t=-\frac{75}{\frac{5}{4}}$$

Divide -20 by \frac{5}{4} by multiplying -20 by the reciprocal of \frac{5}{4}.

$$t^{2}-16t=-\frac{75}{\frac{5}{4}}$$

Divide -75 by \frac{5}{4} by multiplying -75 by the reciprocal of \frac{5}{4}.

$$t^{2}-16t=-60$$

Divide -16, the coefficient of the x term, by 2 to get -8. Then add the square of -8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-16t+\left(-8\right)^{2}=-60+\left(-8\right)^{2}$$

Square -8.

$$t^{2}-16t+64=-60+64$$

Add -60 to 64.

$$t^{2}-16t+64=4$$

Factor t^{2}-16t+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$t-8=2 t-8=-2$$

Add 8 to both sides of the equation.

$$t=10 t=6$$