(5y+9)(3y+1)=0

Solution for (5y+9)(3y+1)=0 equation:

(5y+9)(3y+1)=0

We multiply parentheses ..

(+15y^2+5y+27y+9)=0

We get rid of parentheses

15y^2+5y+27y+9=0

We add all the numbers together, and all the variables

15y^2+32y+9=0

a = 15; b = 32; c = +9;
Δ = b2-4ac
Δ = 322-4·15·9
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-22}{2*15}=\frac{-54}{30}
=-1+4/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+22}{2*15}=\frac{-10}{30}
=-1/3
$