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

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

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

We multiply parentheses ..

(+5y^2-10y+y-2)=0

We get rid of parentheses

5y^2-10y+y-2=0

We add all the numbers together, and all the variables

5y^2-9y-2=0

a = 5; b = -9; c = -2;
Δ = b2-4ac
Δ = -92-4·5·(-2)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-11}{2*5}=\frac{-2}{10}
=-1/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+11}{2*5}=\frac{20}{10}
=2
$