Solve t^2-t=36+4t

$$t ^ { 2 } - t = 36 + 4 t$$

Solve for t

$$t=-4$$

$$t=9$$

Steps Using Factoring

Subtract 36 from both sides.

$$t^{2}-t-36=4t$$

Subtract 4t from both sides.

$$t^{2}-t-36-4t=0$$

Combine -t and -4t to get -5t.

$$t^{2}-5t-36=0$$

To solve the equation, factor t^{2}-5t-36 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.

$$a+b=-5 ab=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -5.

$$a=-9 b=4$$

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

$$\left(t-9\right)\left(t+4\right)$$

To find equation solutions, solve t-9=0 and t+4=0.

$$t=9 t=-4$$

Steps Using Factoring By Grouping

Subtract 36 from both sides.

$$t^{2}-t-36=4t$$

Subtract 4t from both sides.

$$t^{2}-t-36-4t=0$$

Combine -t and -4t to get -5t.

$$t^{2}-5t-36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-36. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-36\right)=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -5.

$$a=-9 b=4$$

Rewrite t^{2}-5t-36 as \left(t^{2}-9t\right)+\left(4t-36\right).

$$\left(t^{2}-9t\right)+\left(4t-36\right)$$

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

$$t\left(t-9\right)+4\left(t-9\right)$$

Factor out common term t-9 by using distributive property.

$$\left(t-9\right)\left(t+4\right)$$

To find equation solutions, solve t-9=0 and t+4=0.

$$t=9 t=-4$$

Steps Using the Quadratic Formula

Subtract 36 from both sides.

$$t^{2}-t-36=4t$$

Subtract 4t from both sides.

$$t^{2}-t-36-4t=0$$

Combine -t and -4t to get -5t.

$$t^{2}-5t-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-36\right)}}{2}$$

Square -5.

$$t=\frac{-\left(-5\right)±\sqrt{25-4\left(-36\right)}}{2}$$

Multiply -4 times -36.

$$t=\frac{-\left(-5\right)±\sqrt{25+144}}{2}$$

Add 25 to 144.

$$t=\frac{-\left(-5\right)±\sqrt{169}}{2}$$

Take the square root of 169.

$$t=\frac{-\left(-5\right)±13}{2}$$

The opposite of -5 is 5.

$$t=\frac{5±13}{2}$$

Now solve the equation t=\frac{5±13}{2} when ± is plus. Add 5 to 13.

$$t=\frac{18}{2}$$

Divide 18 by 2.

$$t=9$$

Now solve the equation t=\frac{5±13}{2} when ± is minus. Subtract 13 from 5.

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

Divide -8 by 2.

$$t=-4$$

The equation is now solved.

$$t=9 t=-4$$

Steps for Completing the Square

Subtract 4t from both sides.

$$t^{2}-t-4t=36$$

Combine -t and -4t to get -5t.

$$t^{2}-5t=36$$

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.

$$t^{2}-5t+\left(-\frac{5}{2}\right)^{2}=36+\left(-\frac{5}{2}\right)^{2}$$

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

$$t^{2}-5t+\frac{25}{4}=36+\frac{25}{4}$$

Add 36 to \frac{25}{4}.

$$t^{2}-5t+\frac{25}{4}=\frac{169}{4}$$

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}.

$$\left(t-\frac{5}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{5}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$t-\frac{5}{2}=\frac{13}{2} t-\frac{5}{2}=-\frac{13}{2}$$

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

$$t=9 t=-4$$