Solve 6t^2=33t+18

$$6 t ^ { 2 } = 33 t + 18$$

Solve for t

$$t=-\frac{1}{2}=-0.5$$

$$t=6$$

Steps Using Factoring By Grouping

Subtract 33t from both sides.

$$6t^{2}-33t=18$$

Subtract 18 from both sides.

$$6t^{2}-33t-18=0$$

Divide both sides by 3.

$$2t^{2}-11t-6=0$$

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

$$a+b=-11 ab=2\left(-6\right)=-12$$

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 -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 -11.

$$a=-12 b=1$$

Rewrite 2t^{2}-11t-6 as \left(2t^{2}-12t\right)+\left(t-6\right).

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

Factor out 2t in 2t^{2}-12t.

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

Factor out common term t-6 by using distributive property.

$$\left(t-6\right)\left(2t+1\right)$$

To find equation solutions, solve t-6=0 and 2t+1=0.

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

Steps Using the Quadratic Formula

Subtract 33t from both sides.

$$6t^{2}-33t=18$$

Subtract 18 from both sides.

$$6t^{2}-33t-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -33 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 6\left(-18\right)}}{2\times 6}$$

Square -33.

$$t=\frac{-\left(-33\right)±\sqrt{1089-4\times 6\left(-18\right)}}{2\times 6}$$

Multiply -4 times 6.

$$t=\frac{-\left(-33\right)±\sqrt{1089-24\left(-18\right)}}{2\times 6}$$

Multiply -24 times -18.

$$t=\frac{-\left(-33\right)±\sqrt{1089+432}}{2\times 6}$$

Add 1089 to 432.

$$t=\frac{-\left(-33\right)±\sqrt{1521}}{2\times 6}$$

Take the square root of 1521.

$$t=\frac{-\left(-33\right)±39}{2\times 6}$$

The opposite of -33 is 33.

$$t=\frac{33±39}{2\times 6}$$

Multiply 2 times 6.

$$t=\frac{33±39}{12}$$

Now solve the equation t=\frac{33±39}{12} when ± is plus. Add 33 to 39.

$$t=\frac{72}{12}$$

Divide 72 by 12.

$$t=6$$

Now solve the equation t=\frac{33±39}{12} when ± is minus. Subtract 39 from 33.

$$t=-\frac{6}{12}$$

Reduce the fraction \frac{-6}{12} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 33t from both sides.

$$6t^{2}-33t=18$$

Divide both sides by 6.

$$\frac{6t^{2}-33t}{6}=\frac{18}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$t^{2}+\left(-\frac{33}{6}\right)t=\frac{18}{6}$$

Reduce the fraction \frac{-33}{6} to lowest terms by extracting and canceling out 3.

$$t^{2}-\frac{11}{2}t=\frac{18}{6}$$

Divide 18 by 6.

$$t^{2}-\frac{11}{2}t=3$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{11}{2}t+\left(-\frac{11}{4}\right)^{2}=3+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{11}{2}t+\frac{121}{16}=3+\frac{121}{16}$$

Add 3 to \frac{121}{16}.

$$t^{2}-\frac{11}{2}t+\frac{121}{16}=\frac{169}{16}$$

Factor t^{2}-\frac{11}{2}t+\frac{121}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{11}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{11}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$t-\frac{11}{4}=\frac{13}{4} t-\frac{11}{4}=-\frac{13}{4}$$

Add \frac{11}{4} to both sides of the equation.

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