5n^2+25n+20=0

Solution for 5n^2+25n+20=0 equation:

5n^2+25n+20=0

a = 5; b = 25; c = +20;
Δ = b2-4ac
Δ = 252-4·5·20
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-15}{2*5}=\frac{-40}{10}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+15}{2*5}=\frac{-10}{10}
=-1
$