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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5y^2+2y-3=0

a = 5; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·5·(-3)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*5}=\frac{-10}{10}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*5}=\frac{6}{10}
=3/5
$