Solve 2k^2-6k-8/3k-12=6

$$\frac { 2 k ^ { 2 } - 6 k - 8 } { 3 k - 12 } = 6$$

Solve for k

$$k=8$$

Steps Using Factoring By Grouping

Variable k cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 3\left(k-4\right).

$$2k^{2}-6k-8=18\left(k-4\right)$$

Use the distributive property to multiply 18 by k-4.

$$2k^{2}-6k-8=18k-72$$

Subtract 18k from both sides.

$$2k^{2}-6k-8-18k=-72$$

Combine -6k and -18k to get -24k.

$$2k^{2}-24k-8=-72$$

Add 72 to both sides.

$$2k^{2}-24k-8+72=0$$

Add -8 and 72 to get 64.

$$2k^{2}-24k+64=0$$

Divide both sides by 2.

$$k^{2}-12k+32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk+32. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\times 32=32$$

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 32.

$$-1,-32 -2,-16 -4,-8$$

Calculate the sum for each pair.

$$-1-32=-33 -2-16=-18 -4-8=-12$$

The solution is the pair that gives sum -12.

$$a=-8 b=-4$$

Rewrite k^{2}-12k+32 as \left(k^{2}-8k\right)+\left(-4k+32\right).

$$\left(k^{2}-8k\right)+\left(-4k+32\right)$$

Factor out k in the first and -4 in the second group.

$$k\left(k-8\right)-4\left(k-8\right)$$

Factor out common term k-8 by using distributive property.

$$\left(k-8\right)\left(k-4\right)$$

To find equation solutions, solve k-8=0 and k-4=0.

$$k=8 k=4$$

Variable k cannot be equal to 4.

$$k=8$$

Steps Using the Quadratic Formula

Variable k cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 3\left(k-4\right).

$$2k^{2}-6k-8=18\left(k-4\right)$$

Use the distributive property to multiply 18 by k-4.

$$2k^{2}-6k-8=18k-72$$

Subtract 18k from both sides.

$$2k^{2}-6k-8-18k=-72$$

Combine -6k and -18k to get -24k.

$$2k^{2}-24k-8=-72$$

Add 72 to both sides.

$$2k^{2}-24k-8+72=0$$

Add -8 and 72 to get 64.

$$2k^{2}-24k+64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -24 for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 2\times 64}}{2\times 2}$$

Square -24.

$$k=\frac{-\left(-24\right)±\sqrt{576-4\times 2\times 64}}{2\times 2}$$

Multiply -4 times 2.

$$k=\frac{-\left(-24\right)±\sqrt{576-8\times 64}}{2\times 2}$$

Multiply -8 times 64.

$$k=\frac{-\left(-24\right)±\sqrt{576-512}}{2\times 2}$$

Add 576 to -512.

$$k=\frac{-\left(-24\right)±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$k=\frac{-\left(-24\right)±8}{2\times 2}$$

The opposite of -24 is 24.

$$k=\frac{24±8}{2\times 2}$$

Multiply 2 times 2.

$$k=\frac{24±8}{4}$$

Now solve the equation k=\frac{24±8}{4} when ± is plus. Add 24 to 8.

$$k=\frac{32}{4}$$

Divide 32 by 4.

$$k=8$$

Now solve the equation k=\frac{24±8}{4} when ± is minus. Subtract 8 from 24.

$$k=\frac{16}{4}$$

Divide 16 by 4.

$$k=4$$

The equation is now solved.

$$k=8 k=4$$

Variable k cannot be equal to 4.

$$k=8$$

Steps for Completing the Square

Variable k cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 3\left(k-4\right).

$$2k^{2}-6k-8=18\left(k-4\right)$$

Use the distributive property to multiply 18 by k-4.

$$2k^{2}-6k-8=18k-72$$

Subtract 18k from both sides.

$$2k^{2}-6k-8-18k=-72$$

Combine -6k and -18k to get -24k.

$$2k^{2}-24k-8=-72$$

Add 8 to both sides.

$$2k^{2}-24k=-72+8$$

Add -72 and 8 to get -64.

$$2k^{2}-24k=-64$$

Divide both sides by 2.

$$\frac{2k^{2}-24k}{2}=-\frac{64}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$k^{2}+\left(-\frac{24}{2}\right)k=-\frac{64}{2}$$

Divide -24 by 2.

$$k^{2}-12k=-\frac{64}{2}$$

Divide -64 by 2.

$$k^{2}-12k=-32$$

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.

$$k^{2}-12k+\left(-6\right)^{2}=-32+\left(-6\right)^{2}$$

Square -6.

$$k^{2}-12k+36=-32+36$$

Add -32 to 36.

$$k^{2}-12k+36=4$$

Factor k^{2}-12k+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}.

$$\left(k-6\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-6\right)^{2}}=\sqrt{4}$$

Simplify.

$$k-6=2 k-6=-2$$

Add 6 to both sides of the equation.

$$k=8 k=4$$

Variable k cannot be equal to 4.

$$k=8$$