Solve 6c^2-72=11c

$$6 c ^ { 2 } - 72 = 11 c$$

Solve for c

$$c = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

$$c = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

Subtract 11c from both sides.

$$6c^{2}-72-11c=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$6c^{2}-11c-72=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6c^{2}+ac+bc-72. To find a and b, set up a system to be solved.

$$a+b=-11 ab=6\left(-72\right)=-432$$

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 -432.

$$1,-432 2,-216 3,-144 4,-108 6,-72 8,-54 9,-48 12,-36 16,-27 18,-24$$

Calculate the sum for each pair.

$$1-432=-431 2-216=-214 3-144=-141 4-108=-104 6-72=-66 8-54=-46 9-48=-39 12-36=-24 16-27=-11 18-24=-6$$

The solution is the pair that gives sum -11.

$$a=-27 b=16$$

Rewrite 6c^{2}-11c-72 as \left(6c^{2}-27c\right)+\left(16c-72\right).

$$\left(6c^{2}-27c\right)+\left(16c-72\right)$$

Factor out 3c in the first and 8 in the second group.

$$3c\left(2c-9\right)+8\left(2c-9\right)$$

Factor out common term 2c-9 by using distributive property.

$$\left(2c-9\right)\left(3c+8\right)$$

To find equation solutions, solve 2c-9=0 and 3c+8=0.

$$c=\frac{9}{2} c=-\frac{8}{3}$$

Steps Using the Quadratic Formula

Subtract 11c from both sides.

$$6c^{2}-72-11c=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$6c^{2}-11c-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -11 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 6\left(-72\right)}}{2\times 6}$$

Square -11.

$$c=\frac{-\left(-11\right)±\sqrt{121-4\times 6\left(-72\right)}}{2\times 6}$$

Multiply -4 times 6.

$$c=\frac{-\left(-11\right)±\sqrt{121-24\left(-72\right)}}{2\times 6}$$

Multiply -24 times -72.

$$c=\frac{-\left(-11\right)±\sqrt{121+1728}}{2\times 6}$$

Add 121 to 1728.

$$c=\frac{-\left(-11\right)±\sqrt{1849}}{2\times 6}$$

Take the square root of 1849.

$$c=\frac{-\left(-11\right)±43}{2\times 6}$$

The opposite of -11 is 11.

$$c=\frac{11±43}{2\times 6}$$

Multiply 2 times 6.

$$c=\frac{11±43}{12}$$

Now solve the equation c=\frac{11±43}{12} when ± is plus. Add 11 to 43.

$$c=\frac{54}{12}$$

Reduce the fraction \frac{54}{12} to lowest terms by extracting and canceling out 6.

$$c=\frac{9}{2}$$

Now solve the equation c=\frac{11±43}{12} when ± is minus. Subtract 43 from 11.

$$c=-\frac{32}{12}$$

Reduce the fraction \frac{-32}{12} to lowest terms by extracting and canceling out 4.

$$c=-\frac{8}{3}$$

The equation is now solved.

$$c=\frac{9}{2} c=-\frac{8}{3}$$

Steps for Completing the Square

Subtract 11c from both sides.

$$6c^{2}-72-11c=0$$

Add 72 to both sides. Anything plus zero gives itself.

$$6c^{2}-11c=72$$

Divide both sides by 6.

$$\frac{6c^{2}-11c}{6}=\frac{72}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$c^{2}-\frac{11}{6}c=\frac{72}{6}$$

Divide 72 by 6.

$$c^{2}-\frac{11}{6}c=12$$

Divide -\frac{11}{6}, the coefficient of the x term, by 2 to get -\frac{11}{12}. Then add the square of -\frac{11}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}-\frac{11}{6}c+\left(-\frac{11}{12}\right)^{2}=12+\left(-\frac{11}{12}\right)^{2}$$

Square -\frac{11}{12} by squaring both the numerator and the denominator of the fraction.

$$c^{2}-\frac{11}{6}c+\frac{121}{144}=12+\frac{121}{144}$$

Add 12 to \frac{121}{144}.

$$c^{2}-\frac{11}{6}c+\frac{121}{144}=\frac{1849}{144}$$

Factor c^{2}-\frac{11}{6}c+\frac{121}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(c-\frac{11}{12}\right)^{2}=\frac{1849}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c-\frac{11}{12}\right)^{2}}=\sqrt{\frac{1849}{144}}$$

Simplify.

$$c-\frac{11}{12}=\frac{43}{12} c-\frac{11}{12}=-\frac{43}{12}$$

Add \frac{11}{12} to both sides of the equation.

$$c=\frac{9}{2} c=-\frac{8}{3}$$