5y^2+40y+25=0

Solution for 5y^2+40y+25=0 equation:

5y^2+40y+25=0

a = 5; b = 40; c = +25;
Δ = b2-4ac
Δ = 402-4·5·25
Δ = 1100
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:

$\sqrt{\Delta}=\sqrt{1100}=\sqrt{100*11}=\sqrt{100}*\sqrt{11}=10\sqrt{11}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{11}}{2*5}=\frac{-40-10\sqrt{11}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{11}}{2*5}=\frac{-40+10\sqrt{11}}{10}
$