(2b+3)(3b-18)=81

Solution for (2b+3)(3b-18)=81 equation:

(2b+3)(3b-18)=81

We move all terms to the left:

(2b+3)(3b-18)-(81)=0

We multiply parentheses ..

(+6b^2-36b+9b-54)-81=0

We get rid of parentheses

6b^2-36b+9b-54-81=0

We add all the numbers together, and all the variables

6b^2-27b-135=0

a = 6; b = -27; c = -135;
Δ = b2-4ac
Δ = -272-4·6·(-135)
Δ = 3969
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3969}=63$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-63}{2*6}=\frac{-36}{12}
=-3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+63}{2*6}=\frac{90}{12}
=7+1/2
$