9b(3b-2)=16

Solution for 9b(3b-2)=16 equation:

9b(3b-2)=16

We move all terms to the left:

9b(3b-2)-(16)=0

We multiply parentheses

27b^2-18b-16=0

a = 27; b = -18; c = -16;
Δ = b2-4ac
Δ = -182-4·27·(-16)
Δ = 2052
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:

$\sqrt{\Delta}=\sqrt{2052}=\sqrt{36*57}=\sqrt{36}*\sqrt{57}=6\sqrt{57}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{57}}{2*27}=\frac{18-6\sqrt{57}}{54}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{57}}{2*27}=\frac{18+6\sqrt{57}}{54}
$