Answer
Solution set: $\{1 + \sqrt7, 1 - \sqrt{7}\}$
Work Step by Step
1. Multiply both sides by 6, to eliminate the fractions.
$(\frac{1}{6}x^2 - \frac{1}{3}x - 1 ) \times 6 =( 0 ) \times 6$
$x^2 - 2x - 6 = 0$
2. Now, use the quadratic formula to find the values for x:
$x = \frac{-(b) ± \sqrt{(b)^2 - 4(a)(c)}}{2(a)}$
$x = \frac{-(-2) ± \sqrt{(-2)^2 - 4(1)(-6)}}{2(1)}$
$x = 1 ± \frac{\sqrt{28}}{2} = 1 ± \frac{\sqrt{28}}{\sqrt{4}}= 1 \sqrt{7}$
Therefore, the solution set is: $\{1 + \sqrt7, 1 - \sqrt{7}\}$
