b(9b+1)=8

Solution for b(9b+1)=8 equation:

b(9b+1)=8

We move all terms to the left:

b(9b+1)-(8)=0

We multiply parentheses

9b^2+b-8=0

a = 9; b = 1; c = -8;
Δ = b2-4ac
Δ = 12-4·9·(-8)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*9}=\frac{-18}{18}
=-1
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*9}=\frac{16}{18}
=8/9
$