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

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

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

We multiply parentheses ..

(+2y^2-4y+y-2)=0

We get rid of parentheses

2y^2-4y+y-2=0

We add all the numbers together, and all the variables

2y^2-3y-2=0

a = 2; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·2·(-2)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*2}=\frac{-2}{4}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*2}=\frac{8}{4}
=2
$