Question

$5x ^{2} + 29x + 20 = 0$what are the real solutions of the equation

Answer 1

$x_1=-5,x_2=-\frac{4}{5}$

1) Considering that the equation is:

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

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

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

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