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

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

=(2A+1)(A+3)

We move all terms to the left:

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

We multiply parentheses ..

-((+2A^2+6A+A+3))=0


We calculate terms in parentheses: -((+2A^2+6A+A+3)), so:

(+2A^2+6A+A+3)

We get rid of parentheses

2A^2+6A+A+3

We add all the numbers together, and all the variables

2A^2+7A+3

Back to the equation:

-(2A^2+7A+3)


We get rid of parentheses

-2A^2-7A-3=0

a = -2; b = -7; c = -3;
Δ = b2-4ac
Δ = -72-4·(-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:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-5}{2*-2}=\frac{2}{-4}
=-1/2
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+5}{2*-2}=\frac{12}{-4}
=-3
$