Answer
{$-\frac{3}{4}\pm\frac{\sqrt {23}}{4}i$}
Work Step by Step
Step 1: $x(2x+3)=-4$
Step 2: $2x^{2}+3x=-4$
Step 3: $2x^{2}+3x+4=0$
Step 4: Comparing $2x^{2}+3x+4=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$;
$a=2$, $b=3$ and $c=4$
Step 5: The quadratic formula is:
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 6: Substituting the values of a,b and c in the formula:
$x=\frac{-(3) \pm \sqrt {(3)^{2}-4(2)(4)}}{2(2)}$
Step 7: $x=\frac{-3 \pm \sqrt {9-32}}{4}$
Step 8: $x=\frac{-3 \pm \sqrt {-23}}{4}$
Step 9: $x=\frac{-3 \pm \sqrt {-1\times23}}{4}$
Step 10: $x=\frac{-3 \pm \sqrt {-1}\times\sqrt {23}}{4}$
Step 11: $x=\frac{-3 \pm i\times\sqrt {23}}{4}$
Step 12: $x=\frac{-3 \pm i\sqrt {23}}{4}$
Step 13: $x=-\frac{3}{4}\pm\frac{\sqrt {23}}{4}i$
Step 14: Therefore, the solution set is {$-\frac{3}{4}\pm\frac{\sqrt {23}}{4}i$}.
