Solve 8=-1/6(x-6)^2+10

$$8 = - \frac { 1 } { 6 } ( x - 6 ) ^ { 2 } + 10$$

Solve for x

$$x=2\sqrt{3}+6\approx 9.464101615$$

$$x=6-2\sqrt{3}\approx 2.535898385$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$8=-\frac{1}{6}\left(x^{2}-12x+36\right)+10$$

Use the distributive property to multiply -\frac{1}{6} by x^{2}-12x+36.

$$8=-\frac{1}{6}x^{2}+2x-6+10$$

Add -6 and 10 to get 4.

$$8=-\frac{1}{6}x^{2}+2x+4$$

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

$$-\frac{1}{6}x^{2}+2x+4=8$$

Subtract 8 from both sides.

$$-\frac{1}{6}x^{2}+2x+4-8=0$$

Subtract 8 from 4 to get -4.

$$-\frac{1}{6}x^{2}+2x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{6} for a, 2 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-\frac{1}{6}\right)\left(-4\right)}}{2\left(-\frac{1}{6}\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-\frac{1}{6}\right)\left(-4\right)}}{2\left(-\frac{1}{6}\right)}$$

Multiply -4 times -\frac{1}{6}.

$$x=\frac{-2±\sqrt{4+\frac{2}{3}\left(-4\right)}}{2\left(-\frac{1}{6}\right)}$$

Multiply \frac{2}{3} times -4.

$$x=\frac{-2±\sqrt{4-\frac{8}{3}}}{2\left(-\frac{1}{6}\right)}$$

Add 4 to -\frac{8}{3}.

$$x=\frac{-2±\sqrt{\frac{4}{3}}}{2\left(-\frac{1}{6}\right)}$$

Take the square root of \frac{4}{3}.

$$x=\frac{-2±\frac{2\sqrt{3}}{3}}{2\left(-\frac{1}{6}\right)}$$

Multiply 2 times -\frac{1}{6}.

$$x=\frac{-2±\frac{2\sqrt{3}}{3}}{-\frac{1}{3}}$$

Now solve the equation x=\frac{-2±\frac{2\sqrt{3}}{3}}{-\frac{1}{3}} when ± is plus. Add -2 to \frac{2\sqrt{3}}{3}.

$$x=\frac{\frac{2\sqrt{3}}{3}-2}{-\frac{1}{3}}$$

Divide -2+\frac{2\sqrt{3}}{3} by -\frac{1}{3} by multiplying -2+\frac{2\sqrt{3}}{3} by the reciprocal of -\frac{1}{3}.

$$x=6-2\sqrt{3}$$

Now solve the equation x=\frac{-2±\frac{2\sqrt{3}}{3}}{-\frac{1}{3}} when ± is minus. Subtract \frac{2\sqrt{3}}{3} from -2.

$$x=\frac{-\frac{2\sqrt{3}}{3}-2}{-\frac{1}{3}}$$

Divide -2-\frac{2\sqrt{3}}{3} by -\frac{1}{3} by multiplying -2-\frac{2\sqrt{3}}{3} by the reciprocal of -\frac{1}{3}.

$$x=2\sqrt{3}+6$$

The equation is now solved.

$$x=6-2\sqrt{3} x=2\sqrt{3}+6$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$8=-\frac{1}{6}\left(x^{2}-12x+36\right)+10$$

Use the distributive property to multiply -\frac{1}{6} by x^{2}-12x+36.

$$8=-\frac{1}{6}x^{2}+2x-6+10$$

Add -6 and 10 to get 4.

$$8=-\frac{1}{6}x^{2}+2x+4$$

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

$$-\frac{1}{6}x^{2}+2x+4=8$$

Subtract 4 from both sides.

$$-\frac{1}{6}x^{2}+2x=8-4$$

Subtract 4 from 8 to get 4.

$$-\frac{1}{6}x^{2}+2x=4$$

Multiply both sides by -6.

$$\frac{-\frac{1}{6}x^{2}+2x}{-\frac{1}{6}}=\frac{4}{-\frac{1}{6}}$$

Dividing by -\frac{1}{6} undoes the multiplication by -\frac{1}{6}.

$$x^{2}+\frac{2}{-\frac{1}{6}}x=\frac{4}{-\frac{1}{6}}$$

Divide 2 by -\frac{1}{6} by multiplying 2 by the reciprocal of -\frac{1}{6}.

$$x^{2}-12x=\frac{4}{-\frac{1}{6}}$$

Divide 4 by -\frac{1}{6} by multiplying 4 by the reciprocal of -\frac{1}{6}.

$$x^{2}-12x=-24$$

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.

$$x^{2}-12x+\left(-6\right)^{2}=-24+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=-24+36$$

Add -24 to 36.

$$x^{2}-12x+36=12$$

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

$$\left(x-6\right)^{2}=12$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{12}$$

Simplify.

$$x-6=2\sqrt{3} x-6=-2\sqrt{3}$$

Add 6 to both sides of the equation.

$$x=2\sqrt{3}+6 x=6-2\sqrt{3}$$