Answer
{$\frac{-1 - i\sqrt {47}}{8},\frac{-1 + i\sqrt {47}}{8}$}
Work Step by Step
Step 1: First, we need to transform $x(4x+1)=-3$ into the standard form of a quadratic equation: $ax^{2}+bx+c=0$:
$x(4x+1)=-3$
$4x^{2}+x+3=0$
Comparing $4x^{2}+x+3=0$ to the standard form of the quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=4$, $b=1$ 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:
$x=\frac{-(1) \pm \sqrt {(1)^{2}-4(4)(3)}}{2(4)}$
Step 4: $x=\frac{-1 \pm \sqrt {1-48}}{8}$
Step 5: $x=\frac{-1 \pm \sqrt {-47}}{8}$
Step 6: $x=\frac{-1 \pm \sqrt {-1\times47}}{8}$
Step 7: $x=\frac{-1 \pm (\sqrt {-1}\times\sqrt {47})}{8}$
Step 8: $x=\frac{-1 \pm (i\times \sqrt {47})}{8}$
Step 9: $x=\frac{-1 \pm i\sqrt {47}}{8}$
Step 10: $x=\frac{-1 - i\sqrt {47}}{8}$ or $x=\frac{-1 + i\sqrt {47}}{8}$
Step 11: Therefore, the solution set is {$\frac{-1 - i\sqrt {47}}{8},\frac{-1 + i\sqrt {47}}{8}$}.
