Solve 0=5t-600/t-10

$$0 = 5 t - \frac { 600 } { t } - 10$$

Solve for t

$$t=-10$$

$$t=12$$

Steps Using Factoring By Grouping

Variable t cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by t.

$$0=5tt-600+t\left(-10\right)$$

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

$$0=5t^{2}-600+t\left(-10\right)$$

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

$$5t^{2}-600+t\left(-10\right)=0$$

Divide both sides by 5.

$$t^{2}-120-2t=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$t^{2}-2t-120=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-120. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-120\right)=-120$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -120.

$$1,-120 2,-60 3,-40 4,-30 5,-24 6,-20 8,-15 10,-12$$

Calculate the sum for each pair.

$$1-120=-119 2-60=-58 3-40=-37 4-30=-26 5-24=-19 6-20=-14 8-15=-7 10-12=-2$$

The solution is the pair that gives sum -2.

$$a=-12 b=10$$

Rewrite t^{2}-2t-120 as \left(t^{2}-12t\right)+\left(10t-120\right).

$$\left(t^{2}-12t\right)+\left(10t-120\right)$$

Factor out t in the first and 10 in the second group.

$$t\left(t-12\right)+10\left(t-12\right)$$

Factor out common term t-12 by using distributive property.

$$\left(t-12\right)\left(t+10\right)$$

To find equation solutions, solve t-12=0 and t+10=0.

$$t=12 t=-10$$

Steps Using the Quadratic Formula

Variable t cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by t.

$$0=5tt-600+t\left(-10\right)$$

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

$$0=5t^{2}-600+t\left(-10\right)$$

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

$$5t^{2}-600+t\left(-10\right)=0$$

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.

$$5t^{2}-10t-600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -10 for b, and -600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 5\left(-600\right)}}{2\times 5}$$

Square -10.

$$t=\frac{-\left(-10\right)±\sqrt{100-4\times 5\left(-600\right)}}{2\times 5}$$

Multiply -4 times 5.

$$t=\frac{-\left(-10\right)±\sqrt{100-20\left(-600\right)}}{2\times 5}$$

Multiply -20 times -600.

$$t=\frac{-\left(-10\right)±\sqrt{100+12000}}{2\times 5}$$

Add 100 to 12000.

$$t=\frac{-\left(-10\right)±\sqrt{12100}}{2\times 5}$$

Take the square root of 12100.

$$t=\frac{-\left(-10\right)±110}{2\times 5}$$

The opposite of -10 is 10.

$$t=\frac{10±110}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{10±110}{10}$$

Now solve the equation t=\frac{10±110}{10} when ± is plus. Add 10 to 110.

$$t=\frac{120}{10}$$

Divide 120 by 10.

$$t=12$$

Now solve the equation t=\frac{10±110}{10} when ± is minus. Subtract 110 from 10.

$$t=-\frac{100}{10}$$

Divide -100 by 10.

$$t=-10$$

The equation is now solved.

$$t=12 t=-10$$

Steps for Completing the Square

Variable t cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by t.

$$0=5tt-600+t\left(-10\right)$$

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

$$0=5t^{2}-600+t\left(-10\right)$$

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

$$5t^{2}-600+t\left(-10\right)=0$$

Add 600 to both sides. Anything plus zero gives itself.

$$5t^{2}+t\left(-10\right)=600$$

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.

$$5t^{2}-10t=600$$

Divide both sides by 5.

$$\frac{5t^{2}-10t}{5}=\frac{600}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$t^{2}+\left(-\frac{10}{5}\right)t=\frac{600}{5}$$

Divide -10 by 5.

$$t^{2}-2t=\frac{600}{5}$$

Divide 600 by 5.

$$t^{2}-2t=120$$

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

$$t^{2}-2t+1=120+1$$

Add 120 to 1.

$$t^{2}-2t+1=121$$

Factor t^{2}-2t+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-1\right)^{2}=121$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-1\right)^{2}}=\sqrt{121}$$

Simplify.

$$t-1=11 t-1=-11$$

Add 1 to both sides of the equation.

$$t=12 t=-10$$