(2y-3)(1+y)=0

Solution for (2y-3)(1+y)=0 equation:

(2y-3)(1+y)=0

We add all the numbers together, and all the variables

(2y-3)(y+1)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2y^2-1y-3=0

a = 2; b = -1; c = -3;
Δ = b2-4ac
Δ = -12-4·2·(-3)
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{-(-1)-5}{2*2}=\frac{-4}{4}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*2}=\frac{6}{4}
=1+1/2
$