Solve 0=3t^2-36t+81

$$0 = 3 { t }^{ 2 } -36t+81$$

Solve for t

$$t=3$$

$$t=9$$

Steps Using Factoring By Grouping

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

$$3t^{2}-36t+81=0$$

Divide both sides by 3.

$$t^{2}-12t+27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt+27. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\times 27=27$$

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 27.

$$-1,-27 -3,-9$$

Calculate the sum for each pair.

$$-1-27=-28 -3-9=-12$$

The solution is the pair that gives sum -12.

$$a=-9 b=-3$$

Rewrite t^{2}-12t+27 as \left(t^{2}-9t\right)+\left(-3t+27\right).

$$\left(t^{2}-9t\right)+\left(-3t+27\right)$$

Factor out t in the first and -3 in the second group.

$$t\left(t-9\right)-3\left(t-9\right)$$

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

$$\left(t-9\right)\left(t-3\right)$$

To find equation solutions, solve t-9=0 and t-3=0.

$$t=9 t=3$$

Steps Using the Quadratic Formula

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

$$3t^{2}-36t+81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -36 for b, and 81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 3\times 81}}{2\times 3}$$

Square -36.

$$t=\frac{-\left(-36\right)±\sqrt{1296-4\times 3\times 81}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-\left(-36\right)±\sqrt{1296-12\times 81}}{2\times 3}$$

Multiply -12 times 81.

$$t=\frac{-\left(-36\right)±\sqrt{1296-972}}{2\times 3}$$

Add 1296 to -972.

$$t=\frac{-\left(-36\right)±\sqrt{324}}{2\times 3}$$

Take the square root of 324.

$$t=\frac{-\left(-36\right)±18}{2\times 3}$$

The opposite of -36 is 36.

$$t=\frac{36±18}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{36±18}{6}$$

Now solve the equation t=\frac{36±18}{6} when ± is plus. Add 36 to 18.

$$t=\frac{54}{6}$$

Divide 54 by 6.

$$t=9$$

Now solve the equation t=\frac{36±18}{6} when ± is minus. Subtract 18 from 36.

$$t=\frac{18}{6}$$

Divide 18 by 6.

$$t=3$$

The equation is now solved.

$$t=9 t=3$$

Steps for Completing the Square

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

$$3t^{2}-36t+81=0$$

Subtract 81 from both sides. Anything subtracted from zero gives its negation.

$$3t^{2}-36t=-81$$

Divide both sides by 3.

$$\frac{3t^{2}-36t}{3}=-\frac{81}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$t^{2}+\left(-\frac{36}{3}\right)t=-\frac{81}{3}$$

Divide -36 by 3.

$$t^{2}-12t=-\frac{81}{3}$$

Divide -81 by 3.

$$t^{2}-12t=-27$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-12t+\left(-6\right)^{2}=-27+\left(-6\right)^{2}$$

Square -6.

$$t^{2}-12t+36=-27+36$$

Add -27 to 36.

$$t^{2}-12t+36=9$$

Factor t^{2}-12t+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-6\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-6\right)^{2}}=\sqrt{9}$$

Simplify.

$$t-6=3 t-6=-3$$

Add 6 to both sides of the equation.

$$t=9 t=3$$