Answer
{$\frac{-3 - 5i\sqrt {3}}{14},\frac{-3 + 5i\sqrt {3}}{14}$}
Work Step by Step
Step 1: Comparing $7x^{2}+3x+3=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=7$, $b=3$ and $c=3$
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:
$x=\frac{-(3) \pm \sqrt {(3)^{2}-4(7)(3)}}{2(7)}$
Step 4: $x=\frac{-3 \pm \sqrt {9-84}}{14}$
Step 5: $x=\frac{-3 \pm \sqrt {-75}}{14}$
Step 6: $x=\frac{-3 \pm \sqrt {-1\times75}}{14}$
Step 7: $x=\frac{-3 \pm (\sqrt {-1}\times\sqrt {25\times3})}{14}$
Step 8: $x=\frac{-3 \pm (i\times 5\sqrt {3})}{14}$
Step 9: $x=\frac{-3 \pm i5\sqrt {3}}{14}$
Step 10: $x=\frac{-3 - 5i\sqrt {3}}{14}$ or $x=\frac{-3 + 5i\sqrt {3}}{14}$
Step 11: Therefore, the solution set is {$\frac{-3 - 5i\sqrt {3}}{14},\frac{-3 + 5i\sqrt {3}}{14}$}.
