Answer
{$-\frac{2}{3},\frac{1}{2}$}
Work Step by Step
Step 1: Comparing $6x^{2}+x-2=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=6$, $b=1$, and $c=-2$
Step 2: The quadratic formula is:
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a,b, and c in the formula, we obtain:
$x=\frac{-(1) \pm \sqrt {(1)^{2}-4(6)(-2)}}{2(6)}$
Step 4: $x=\frac{-1 \pm \sqrt {1+48}}{12}$
Step 5: $x=\frac{-1 \pm \sqrt {49}}{12}$
Step 6: $x=\frac{-1 \pm 7}{12}$
Step 7: $x=\frac{-1 + 7}{12}$ or $x=\frac{-1 - 7}{12}$
Step 8: $x=\frac{6}{12}$ or $x=\frac{-8}{12}$
Step 9: $x=\frac{1}{2}$ or $x=\frac{-2}{3}$
Step 10: Therefore, the solution set is {$-\frac{2}{3},\frac{1}{2}$}.
