Solve 6/2t^2+2t=t-2/t+1

$$\frac { 6 } { 2 t ^ { 2 } + 2 t } = \frac { t - 2 } { t + 1 }$$

Solve for t

$$t=3$$

Steps Using Factoring By Grouping

Variable t cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by 2t\left(t+1\right), the least common multiple of 2t^{2}+2t,t+1.

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

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

$$6=2t^{2}-4t$$

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

$$2t^{2}-4t=6$$

Subtract 6 from both sides.

$$2t^{2}-4t-6=0$$

Divide both sides by 2.

$$t^{2}-2t-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-3\right)=-3$$

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. The only such pair is the system solution.

$$a=-3 b=1$$

Rewrite t^{2}-2t-3 as \left(t^{2}-3t\right)+\left(t-3\right).

$$\left(t^{2}-3t\right)+\left(t-3\right)$$

Factor out t in t^{2}-3t.

$$t\left(t-3\right)+t-3$$

Factor out common term t-3 by using distributive property.

$$\left(t-3\right)\left(t+1\right)$$

To find equation solutions, solve t-3=0 and t+1=0.

$$t=3 t=-1$$

Variable t cannot be equal to -1.

$$t=3$$

Steps Using the Quadratic Formula

Variable t cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by 2t\left(t+1\right), the least common multiple of 2t^{2}+2t,t+1.

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

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

$$6=2t^{2}-4t$$

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

$$2t^{2}-4t=6$$

Subtract 6 from both sides.

$$2t^{2}-4t-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Square -4.

$$t=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$t=\frac{-\left(-4\right)±\sqrt{16-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$t=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 2}$$

Add 16 to 48.

$$t=\frac{-\left(-4\right)±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$t=\frac{-\left(-4\right)±8}{2\times 2}$$

The opposite of -4 is 4.

$$t=\frac{4±8}{2\times 2}$$

Multiply 2 times 2.

$$t=\frac{4±8}{4}$$

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

$$t=\frac{12}{4}$$

Divide 12 by 4.

$$t=3$$

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

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

Divide -4 by 4.

$$t=-1$$

The equation is now solved.

$$t=3 t=-1$$

Variable t cannot be equal to -1.

$$t=3$$

Steps for Completing the Square

Variable t cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by 2t\left(t+1\right), the least common multiple of 2t^{2}+2t,t+1.

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

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

$$6=2t^{2}-4t$$

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

$$2t^{2}-4t=6$$

Divide both sides by 2.

$$\frac{2t^{2}-4t}{2}=\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

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

Divide -4 by 2.

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

Divide 6 by 2.

$$t^{2}-2t=3$$

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.

$$t^{2}-2t+1=3+1$$

Add 3 to 1.

$$t^{2}-2t+1=4$$

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}.

$$\left(t-1\right)^{2}=4$$

Take the square root of both sides of the equation.

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

Simplify.

$$t-1=2 t-1=-2$$

Add 1 to both sides of the equation.

$$t=3 t=-1$$

Variable t cannot be equal to -1.

$$t=3$$