Question

`5x ^(2) + 29x + 20 = 0`what are the real solutions of the equation

Answer 1

`x_1=-5,x_2=-(4)/(5)`

1) Considering that the equation is:

`5x^2+29x+20=0`

2) Let's solve it, then making use of the quadratic formula:

`\begin{gathered} x=\frac{-b\pm\sqrt[]{\Delta}}{2a} \\ x=\frac{-29\pm\sqrt[]{(29)^2-4(5)(20)}}{2(5)} \\ x=\frac{-29\pm\sqrt[]{441}}{10}=(29\pm21)/(10) \\ x_1=-(4)/(5) \\ x_2=-5 \end{gathered}`

3) Hence, the Solution Set is S= { -5,-4/5}