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