5m^2+35m+20=0

Solution for 5m^2+35m+20=0 equation:

5m^2+35m+20=0

a = 5; b = 35; c = +20;
Δ = b2-4ac
Δ = 352-4·5·20
Δ = 825
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-5\sqrt{33}}{2*5}=\frac{-35-5\sqrt{33}}{10}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+5\sqrt{33}}{2*5}=\frac{-35+5\sqrt{33}}{10}
$