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

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

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

We multiply parentheses ..

(+3y^2+15y-1y-5)=0

We get rid of parentheses

3y^2+15y-1y-5=0

We add all the numbers together, and all the variables

3y^2+14y-5=0

a = 3; b = 14; c = -5;
Δ = b2-4ac
Δ = 142-4·3·(-5)
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{-(14)-16}{2*3}=\frac{-30}{6}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+16}{2*3}=\frac{2}{6}
=1/3
$