(y+3)(2y+3)-2y=18

Solution for (y+3)(2y+3)-2y=18 equation:

(y+3)(2y+3)-2y=18

We move all terms to the left:

(y+3)(2y+3)-2y-(18)=0

We add all the numbers together, and all the variables

-2y+(y+3)(2y+3)-18=0

We multiply parentheses ..

(+2y^2+3y+6y+9)-2y-18=0

We get rid of parentheses

2y^2+3y+6y-2y+9-18=0

We add all the numbers together, and all the variables

2y^2+7y-9=0

a = 2; b = 7; c = -9;
Δ = b2-4ac
Δ = 72-4·2·(-9)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-11}{2*2}=\frac{-18}{4}
=-4+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+11}{2*2}=\frac{4}{4}
=1
$