y^2+50y+25=0

Solution for y^2+50y+25=0 equation:

y^2+50y+25=0

a = 1; b = 50; c = +25;
Δ = b2-4ac
Δ = 502-4·1·25
Δ = 2400
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:

$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-20\sqrt{6}}{2*1}=\frac{-50-20\sqrt{6}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+20\sqrt{6}}{2*1}=\frac{-50+20\sqrt{6}}{2}
$