5y^2+16y=0

Solution for 5y^2+16y=0 equation:

5y^2+16y=0

a = 5; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·5·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*5}=\frac{-32}{10}
=-3+1/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*5}=\frac{0}{10}
=0
$