[tex]2x^2 + 3x = -8\\\\2x^2 + 3x+8=0\\a=2 , \ \ b=3, \ \ c=8\\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-3-\sqrt{3^2-4 \cdot 2\cdot8 }}{2a}=\frac{-3-\sqrt{9-64}}{2 \cdot 2}=\frac{-3-\sqrt{-55}}{4}=\frac{-3-\sqrt{ 55}i}{4} \\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}==\frac{-3+\sqrt{9-64}}{2 \cdot 2}=\frac{-3+\sqrt{-55}}{4}=\frac{-3+\sqrt{ 55}i}{4}\\ \\Answer: \ If b^2-4ac < 0, \ then \ roots \ are \ imaginary \ (non-real) [/tex]
