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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+2y^2+6y+y+3)-1=0

We get rid of parentheses

2y^2+6y+y+3-1=0

We add all the numbers together, and all the variables

2y^2+7y+2=0

a = 2; b = 7; c = +2;
Δ = b2-4ac
Δ = 72-4·2·2
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{33}}{2*2}=\frac{-7-\sqrt{33}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{33}}{2*2}=\frac{-7+\sqrt{33}}{4}
$