-3(b-8)-5=9(b2)+1

Solution for -3(b-8)-5=9(b2)+1 equation:

-3(b-8)-5=9(b2)+1

We move all terms to the left:

-3(b-8)-5-(9(b2)+1)=0

We add all the numbers together, and all the variables

-(+9b^2+1)-3(b-8)-5=0

We multiply parentheses

-(+9b^2+1)-3b+24-5=0

We get rid of parentheses

-9b^2-3b-1+24-5=0

We add all the numbers together, and all the variables

-9b^2-3b+18=0

a = -9; b = -3; c = +18;
Δ = b2-4ac
Δ = -32-4·(-9)·18
Δ = 657
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{657}=\sqrt{9*73}=\sqrt{9}*\sqrt{73}=3\sqrt{73}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{73}}{2*-9}=\frac{3-3\sqrt{73}}{-18}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{73}}{2*-9}=\frac{3+3\sqrt{73}}{-18}
$