y^2+20y-4=0

Solution for y^2+20y-4=0 equation:

y^2+20y-4=0

a = 1; b = 20; c = -4;
Δ = b2-4ac
Δ = 202-4·1·(-4)
Δ = 416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{26}}{2*1}=\frac{-20-4\sqrt{26}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{26}}{2*1}=\frac{-20+4\sqrt{26}}{2}
$