Given Question
[tex]\begin{gathered} 2x^2+5\text{ = -3x} \\ Step1\colon\text{rearrange the eqaution to be in the form of a quadratic equation} \\ 2x^2+5\text{ +3x = 0} \\ 2x^2\text{ +3x +5 = 0} \end{gathered}[/tex]
Step 2
Define and assign values to the quadratic formula given
[tex]y\text{ =}\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex][tex]\begin{gathered} \text{From the general quadratic equation} \\ ax^2+bx+c=0 \\ a\text{ =2} \\ b=\text{ 3} \\ c\text{ =5} \end{gathered}[/tex][tex]\begin{gathered} \text{Step 3} \\ \text{substitute those values into the formula and solve for x} \\ x=\frac{-3\pm\sqrt[]{3^2-4\times2\times5}}{2\times2} \\ x=\frac{-3\pm\sqrt[]{9-40}}{4} \\ x=\frac{-3\pm\sqrt[]{-31}}{4} \\ x=\frac{-3\pm\sqrt[]{-1}\times\sqrt[]{31}}{4} \\ \text{but i = }\sqrt[]{-1},\text{ therefore} \\ x\text{ =}\frac{-3\pm\sqrt[]{31}i}{4} \end{gathered}[/tex][tex]\begin{gathered} x\text{ = }\frac{-3+\sqrt[]{31}i}{4} \\ x\text{ = -}\frac{3}{4}+\frac{\sqrt[]{-31}}{4} \\ or\text{ } \\ x\text{ = }\frac{-3-\sqrt[]{31}i}{4} \\ x\text{ = -}\frac{3}{4}-\frac{\sqrt[]{-31}}{4} \\ \end{gathered}[/tex]
Therefore,
