Answer
The solution set in standard form is $$\Big\{-\frac{3}{4}\pm\frac{\sqrt7}{4}i\Big\}$$
Work Step by Step
$$2x^2+3x=-2$$
First, write the equation in standard form.
$$2x^2+3x+2=0$$
Now use the quadratic formula.
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
As $a=2, b=3, c=2$
$$x=\frac{-3\pm\sqrt{3^2-4\times2\times2}}{2\times2}$$
$$x=\frac{-3\pm\sqrt{9-16}}{4}$$
$$x=\frac{-3\pm\sqrt{-7}}{4}$$
Now we rewrite $\sqrt{-7}=i\sqrt7$
$$x=\frac{-3\pm i\sqrt7}{4}$$
The solution set in standard form is $$\Big\{-\frac{3}{4}\pm\frac{\sqrt7}{4}i\Big\}$$
