3b^2-3b=18

Solution for 3b^2-3b=18 equation:

3b^2-3b=18

We move all terms to the left:

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

a = 3; b = -3; c = -18;
Δ = b2-4ac
Δ = -32-4·3·(-18)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-15}{2*3}=\frac{-12}{6}
=-2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+15}{2*3}=\frac{18}{6}
=3
$