Solve t+3/t^2-2t=10/t^2-4

$$\frac { t + 3 } { t ^ { 2 } - 2 t } = \frac { 10 } { t ^ { 2 } - 4 }$$

Solve for t

$$t=3$$

Steps Using Factoring

Variable t cannot be equal to any of the values -2,0,2 since division by zero is not defined. Multiply both sides of the equation by t\left(t-2\right)\left(t+2\right), the least common multiple of t^{2}-2t,t^{2}-4.

$$\left(t+2\right)\left(t+3\right)=t\times 10$$

Use the distributive property to multiply t+2 by t+3 and combine like terms.

$$t^{2}+5t+6=t\times 10$$

Subtract t\times 10 from both sides.

$$t^{2}+5t+6-t\times 10=0$$

Combine 5t and -t\times 10 to get -5t.

$$t^{2}-5t+6=0$$

To solve the equation, factor t^{2}-5t+6 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.

$$a+b=-5 ab=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

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

$$\left(t-3\right)\left(t-2\right)$$

To find equation solutions, solve t-3=0 and t-2=0.

$$t=3 t=2$$

Variable t cannot be equal to 2.

$$t=3$$

Steps Using Factoring By Grouping

Variable t cannot be equal to any of the values -2,0,2 since division by zero is not defined. Multiply both sides of the equation by t\left(t-2\right)\left(t+2\right), the least common multiple of t^{2}-2t,t^{2}-4.

$$\left(t+2\right)\left(t+3\right)=t\times 10$$

Use the distributive property to multiply t+2 by t+3 and combine like terms.

$$t^{2}+5t+6=t\times 10$$

Subtract t\times 10 from both sides.

$$t^{2}+5t+6-t\times 10=0$$

Combine 5t and -t\times 10 to get -5t.

$$t^{2}-5t+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt+6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\times 6=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

Rewrite t^{2}-5t+6 as \left(t^{2}-3t\right)+\left(-2t+6\right).

$$\left(t^{2}-3t\right)+\left(-2t+6\right)$$

Factor out t in the first and -2 in the second group.

$$t\left(t-3\right)-2\left(t-3\right)$$

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

$$\left(t-3\right)\left(t-2\right)$$

To find equation solutions, solve t-3=0 and t-2=0.

$$t=3 t=2$$

Variable t cannot be equal to 2.

$$t=3$$

Steps Using the Quadratic Formula

Variable t cannot be equal to any of the values -2,0,2 since division by zero is not defined. Multiply both sides of the equation by t\left(t-2\right)\left(t+2\right), the least common multiple of t^{2}-2t,t^{2}-4.

$$\left(t+2\right)\left(t+3\right)=t\times 10$$

Use the distributive property to multiply t+2 by t+3 and combine like terms.

$$t^{2}+5t+6=t\times 10$$

Subtract t\times 10 from both sides.

$$t^{2}+5t+6-t\times 10=0$$

Combine 5t and -t\times 10 to get -5t.

$$t^{2}-5t+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6}}{2}$$

Square -5.

$$t=\frac{-\left(-5\right)±\sqrt{25-4\times 6}}{2}$$

Multiply -4 times 6.

$$t=\frac{-\left(-5\right)±\sqrt{25-24}}{2}$$

Add 25 to -24.

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

Take the square root of 1.

$$t=\frac{-\left(-5\right)±1}{2}$$

The opposite of -5 is 5.

$$t=\frac{5±1}{2}$$

Now solve the equation t=\frac{5±1}{2} when ± is plus. Add 5 to 1.

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

Divide 6 by 2.

$$t=3$$

Now solve the equation t=\frac{5±1}{2} when ± is minus. Subtract 1 from 5.

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

Divide 4 by 2.

$$t=2$$

The equation is now solved.

$$t=3 t=2$$

Variable t cannot be equal to 2.

$$t=3$$

Steps for Completing the Square

Variable t cannot be equal to any of the values -2,0,2 since division by zero is not defined. Multiply both sides of the equation by t\left(t-2\right)\left(t+2\right), the least common multiple of t^{2}-2t,t^{2}-4.

$$\left(t+2\right)\left(t+3\right)=t\times 10$$

Use the distributive property to multiply t+2 by t+3 and combine like terms.

$$t^{2}+5t+6=t\times 10$$

Subtract t\times 10 from both sides.

$$t^{2}+5t+6-t\times 10=0$$

Combine 5t and -t\times 10 to get -5t.

$$t^{2}-5t+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$t^{2}-5t=-6$$

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

$$t^{2}-5t+\left(-\frac{5}{2}\right)^{2}=-6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-5t+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$t^{2}-5t+\frac{25}{4}=\frac{1}{4}$$

Factor t^{2}-5t+\frac{25}{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}.

$$\left(t-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$t-\frac{5}{2}=\frac{1}{2} t-\frac{5}{2}=-\frac{1}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$t=3 t=2$$

Variable t cannot be equal to 2.

$$t=3$$