Answer
{$-3.55,1.22$}
Work Step by Step
Step 1: Comparing $3x^{2}+7x-13=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=3$, $b=7$, and $c=-13$
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:
$x=\frac{-(7) \pm \sqrt {(7)^{2}-4(3)(-13)}}{2(3)}$
Step 4: $x=\frac{-7 \pm \sqrt {49+156}}{6}$
Step 5: $x=\frac{-7 \pm \sqrt {205}}{6}$
Step 6: $x=\frac{-7 \pm 14.318}{6}$
Step 7: $x=\frac{-7+14.318}{6}$ or $x=\frac{-7-14.318}{6}$
Step 8: $x=\frac{7.318}{6}$ or $x=\frac{-21.318}{6}$
Step 9: $x=1.220\approx1.22$ or $x=-3.553\approx-3.55$
Step 10: Therefore, the solution set is {$-3.55,1.22$}.
