Solve 2/t=t/-2t+6

$$\frac { 2 } { t } = \frac { t } { - 2 t + 6 }$$

Solve for t

$$t=-6$$

$$t=2$$

Steps Using Factoring

Variable t cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by 2t\left(t-3\right), the least common multiple of t,-2t+6.

$$\left(2t-6\right)\times 2=-tt$$

Use the distributive property to multiply 2t-6 by 2.

$$4t-12=-tt$$

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

$$4t-12=-t^{2}$$

Add t^{2} to both sides.

$$4t-12+t^{2}=0$$

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

$$t^{2}+4t-12=0$$

To solve the equation, factor t^{2}+4t-12 using formula t^{2}+\left(a+b\right)t+ab=\left(t+a\right)\left(t+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=-2 b=6$$

Rewrite factored expression \left(t+a\right)\left(t+b\right) using the obtained values.

$$\left(t-2\right)\left(t+6\right)$$

To find equation solutions, solve t-2=0 and t+6=0.

$$t=2 t=-6$$

Steps Using Factoring By Grouping

Variable t cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by 2t\left(t-3\right), the least common multiple of t,-2t+6.

$$\left(2t-6\right)\times 2=-tt$$

Use the distributive property to multiply 2t-6 by 2.

$$4t-12=-tt$$

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

$$4t-12=-t^{2}$$

Add t^{2} to both sides.

$$4t-12+t^{2}=0$$

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

$$t^{2}+4t-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-12. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-12\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=-2 b=6$$

Rewrite t^{2}+4t-12 as \left(t^{2}-2t\right)+\left(6t-12\right).

$$\left(t^{2}-2t\right)+\left(6t-12\right)$$

Factor out t in the first and 6 in the second group.

$$t\left(t-2\right)+6\left(t-2\right)$$

Factor out common term t-2 by using distributive property.

$$\left(t-2\right)\left(t+6\right)$$

To find equation solutions, solve t-2=0 and t+6=0.

$$t=2 t=-6$$

Steps Using the Quadratic Formula

Variable t cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by 2t\left(t-3\right), the least common multiple of t,-2t+6.

$$\left(2t-6\right)\times 2=-tt$$

Use the distributive property to multiply 2t-6 by 2.

$$4t-12=-tt$$

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

$$4t-12=-t^{2}$$

Add t^{2} to both sides.

$$4t-12+t^{2}=0$$

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.

$$t^{2}+4t-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-4±\sqrt{4^{2}-4\left(-12\right)}}{2}$$

Square 4.

$$t=\frac{-4±\sqrt{16-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$t=\frac{-4±\sqrt{16+48}}{2}$$

Add 16 to 48.

$$t=\frac{-4±\sqrt{64}}{2}$$

Take the square root of 64.

$$t=\frac{-4±8}{2}$$

Now solve the equation t=\frac{-4±8}{2} when ± is plus. Add -4 to 8.

$$t=\frac{4}{2}$$

Divide 4 by 2.

$$t=2$$

Now solve the equation t=\frac{-4±8}{2} when ± is minus. Subtract 8 from -4.

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

Divide -12 by 2.

$$t=-6$$

The equation is now solved.

$$t=2 t=-6$$

Steps for Completing the Square

Variable t cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by 2t\left(t-3\right), the least common multiple of t,-2t+6.

$$\left(2t-6\right)\times 2=-tt$$

Use the distributive property to multiply 2t-6 by 2.

$$4t-12=-tt$$

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

$$4t-12=-t^{2}$$

Add t^{2} to both sides.

$$4t-12+t^{2}=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$4t+t^{2}=12$$

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.

$$t^{2}+4t=12$$

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

$$t^{2}+4t+2^{2}=12+2^{2}$$

Square 2.

$$t^{2}+4t+4=12+4$$

Add 12 to 4.

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

Factor t^{2}+4t+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(t+2\right)^{2}}=\sqrt{16}$$

Simplify.

$$t+2=4 t+2=-4$$

Subtract 2 from both sides of the equation.

$$t=2 t=-6$$