(1+y)(5y-2)=0

Solution for (1+y)(5y-2)=0 equation:

(1+y)(5y-2)=0

We add all the numbers together, and all the variables

(y+1)(5y-2)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5y^2+3y-2=0

a = 5; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·5·(-2)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*5}=\frac{-10}{10}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*5}=\frac{4}{10}
=2/5
$