3b^2=18b+17=b

Solution for 3b^2=18b+17=b equation:

3b^2=18b+17=b

We move all terms to the left:

3b^2-(18b+17)=0

We get rid of parentheses

3b^2-18b-17=0

a = 3; b = -18; c = -17;
Δ = b2-4ac
Δ = -182-4·3·(-17)
Δ = 528
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:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{33}}{2*3}=\frac{18-4\sqrt{33}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{33}}{2*3}=\frac{18+4\sqrt{33}}{6}
$