Answer
{$-\frac{3}{4}$}
Work Step by Step
Step 1: Comparing $16x^{2}+24x+9=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=16$, $b=24$ and $c=9$
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:
$x=\frac{-(24) \pm \sqrt {(24)^{2}-4(16)(9)}}{2(16)}$
Step 4: $x=\frac{-24 \pm \sqrt {576-576}}{32}$
Step 5: $x=\frac{-24 \pm \sqrt {0}}{32}$
Step 6: $x=\frac{-24 \pm 0}{32}$
Step 7: $x=\frac{-24}{32}$
Step 8: $x=\frac{-3}{4}$
Step 9: Therefore, the solution set is {$-\frac{3}{4}$}.
