Solve 2/t=t/5t-12

$$\frac { 2 } { t } = \frac { t } { 5 t - 12 }$$

Solve for t

$$t=4$$

$$t=6$$

Steps Using Factoring By Grouping

Variable t cannot be equal to any of the values 0,\frac{12}{5} since division by zero is not defined. Multiply both sides of the equation by t\left(5t-12\right), the least common multiple of t,5t-12.

$$\left(5t-12\right)\times 2=tt$$

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

$$\left(5t-12\right)\times 2=t^{2}$$

Use the distributive property to multiply 5t-12 by 2.

$$10t-24=t^{2}$$

Subtract t^{2} from both sides.

$$10t-24-t^{2}=0$$

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

$$-t^{2}+10t-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -t^{2}+at+bt-24. To find a and b, set up a system to be solved.

$$a+b=10 ab=-\left(-24\right)=24$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 10.

$$a=6 b=4$$

Rewrite -t^{2}+10t-24 as \left(-t^{2}+6t\right)+\left(4t-24\right).

$$\left(-t^{2}+6t\right)+\left(4t-24\right)$$

Factor out -t in the first and 4 in the second group.

$$-t\left(t-6\right)+4\left(t-6\right)$$

Factor out common term t-6 by using distributive property.

$$\left(t-6\right)\left(-t+4\right)$$

To find equation solutions, solve t-6=0 and -t+4=0.

$$t=6 t=4$$

Steps Using the Quadratic Formula

Variable t cannot be equal to any of the values 0,\frac{12}{5} since division by zero is not defined. Multiply both sides of the equation by t\left(5t-12\right), the least common multiple of t,5t-12.

$$\left(5t-12\right)\times 2=tt$$

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

$$\left(5t-12\right)\times 2=t^{2}$$

Use the distributive property to multiply 5t-12 by 2.

$$10t-24=t^{2}$$

Subtract t^{2} from both sides.

$$10t-24-t^{2}=0$$

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.

$$-t^{2}+10t-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 10 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-24\right)}}{2\left(-1\right)}$$

Square 10.

$$t=\frac{-10±\sqrt{100-4\left(-1\right)\left(-24\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-10±\sqrt{100+4\left(-24\right)}}{2\left(-1\right)}$$

Multiply 4 times -24.

$$t=\frac{-10±\sqrt{100-96}}{2\left(-1\right)}$$

Add 100 to -96.

$$t=\frac{-10±\sqrt{4}}{2\left(-1\right)}$$

Take the square root of 4.

$$t=\frac{-10±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{-10±2}{-2}$$

Now solve the equation t=\frac{-10±2}{-2} when ± is plus. Add -10 to 2.

$$t=-\frac{8}{-2}$$

Divide -8 by -2.

$$t=4$$

Now solve the equation t=\frac{-10±2}{-2} when ± is minus. Subtract 2 from -10.

$$t=-\frac{12}{-2}$$

Divide -12 by -2.

$$t=6$$

The equation is now solved.

$$t=4 t=6$$

Steps for Completing the Square

Variable t cannot be equal to any of the values 0,\frac{12}{5} since division by zero is not defined. Multiply both sides of the equation by t\left(5t-12\right), the least common multiple of t,5t-12.

$$\left(5t-12\right)\times 2=tt$$

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

$$\left(5t-12\right)\times 2=t^{2}$$

Use the distributive property to multiply 5t-12 by 2.

$$10t-24=t^{2}$$

Subtract t^{2} from both sides.

$$10t-24-t^{2}=0$$

Add 24 to both sides. Anything plus zero gives itself.

$$10t-t^{2}=24$$

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.

$$-t^{2}+10t=24$$

Divide both sides by -1.

$$\frac{-t^{2}+10t}{-1}=\frac{24}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\frac{10}{-1}t=\frac{24}{-1}$$

Divide 10 by -1.

$$t^{2}-10t=\frac{24}{-1}$$

Divide 24 by -1.

$$t^{2}-10t=-24$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-10t+\left(-5\right)^{2}=-24+\left(-5\right)^{2}$$

Square -5.

$$t^{2}-10t+25=-24+25$$

Add -24 to 25.

$$t^{2}-10t+25=1$$

Factor t^{2}-10t+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-5\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-5\right)^{2}}=\sqrt{1}$$

Simplify.

$$t-5=1 t-5=-1$$

Add 5 to both sides of the equation.

$$t=6 t=4$$