2(3b+1)b=12

Solution for 2(3b+1)b=12 equation:

2(3b+1)b=12

We move all terms to the left:

2(3b+1)b-(12)=0

We multiply parentheses

6b^2+2b-12=0

a = 6; b = 2; c = -12;
Δ = b2-4ac
Δ = 22-4·6·(-12)
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{73}}{2*6}=\frac{-2-2\sqrt{73}}{12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{73}}{2*6}=\frac{-2+2\sqrt{73}}{12}
$