16y^2+16y+3y=0

Solution for 16y^2+16y+3y=0 equation:

16y^2+16y+3y=0

We add all the numbers together, and all the variables

16y^2+19y=0

a = 16; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·16·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*16}=\frac{-38}{32}
=-1+3/16
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*16}=\frac{0}{32}
=0
$