We have the equation:
[tex]x^2+3x=40[/tex]We have to find its roots with the quadratic formula:
[tex]\begin{gathered} x^2+3x-40=0 \\ x=\frac{-3}{2\cdot1}\pm\frac{\sqrt[]{3^2-4\cdot1\cdot(-40)}}{2\cdot1} \\ x=-\frac{3}{2}\pm\frac{\sqrt[]{9+160}}{2} \\ x=-\frac{3}{2}\pm\frac{\sqrt[]{169}}{2} \\ x=-\frac{3}{2}\pm\frac{13}{2} \\ x_1=-\frac{3}{2}-\frac{13}{2}=-\frac{16}{2}=-8 \\ x_2=-\frac{3}{2}+\frac{13}{2}=\frac{10}{2}=5 \end{gathered}[/tex]Answer:
The roots (or solutions) of the equation are:
x1 = -8
x2 = 5
