39y^2+16y=0

Solution for 39y^2+16y=0 equation:

39y^2+16y=0

a = 39; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·39·0
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*39}=\frac{-32}{78}
=-16/39
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*39}=\frac{0}{78}
=0
$