Answer
{$-\frac{5}{4},-\frac{1}{3}$}
Work Step by Step
Step 1: Comparing $12x^{2}+19x+5=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=12$, $b=19$, and $c=5$
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:
$x=\frac{-(19) \pm \sqrt {(19)^{2}-4(12)(5)}}{2(12)}$
Step 4: $x=\frac{-19 \pm \sqrt {361-240}}{24}$
Step 5: $x=\frac{-19 \pm \sqrt {121}}{24}$
Step 6: $x=\frac{-19 \pm 11}{24}$
Step 7: $x=\frac{-19+11}{24}$ or $x=\frac{-19-11}{24}$
Step 8: $x=\frac{-8}{24}$ or $x=\frac{-30}{24}$
Step 9: $x=-\frac{1}{3}$ or $x=-\frac{5}{4}$
Step 10: Therefore, the solution set is {$-\frac{5}{4},-\frac{1}{3}$}.
