b^2=9b-8

Solution for b^2=9b-8 equation:

b^2=9b-8

We move all terms to the left:

b^2-(9b-8)=0

We get rid of parentheses

b^2-9b+8=0

a = 1; b = -9; c = +8;
Δ = b2-4ac
Δ = -92-4·1·8
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-7}{2*1}=\frac{2}{2}
=1
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+7}{2*1}=\frac{16}{2}
=8
$