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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2y^2+5y+3=0

a = 2; b = 5; c = +3;
Δ = b2-4ac
Δ = 52-4·2·3
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*2}=\frac{-6}{4}
=-1+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*2}=\frac{-4}{4}
=-1
$