5n^2+20n-5=0

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

5n^2+20n-5=0

a = 5; b = 20; c = -5;
Δ = b2-4ac
Δ = 202-4·5·(-5)
Δ = 500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{5}}{2*5}=\frac{-20-10\sqrt{5}}{10}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{5}}{2*5}=\frac{-20+10\sqrt{5}}{10}
$