-3b(b-1)=-6(1+2b)

Solution for -3b(b-1)=-6(1+2b) equation:

-3b(b-1)=-6(1+2b)

We move all terms to the left:

-3b(b-1)-(-6(1+2b))=0

We add all the numbers together, and all the variables

-3b(b-1)-(-6(2b+1))=0

We multiply parentheses

-3b^2+3b-(-6(2b+1))=0


We calculate terms in parentheses: -(-6(2b+1)), so:

-6(2b+1)

We multiply parentheses

-12b-6

Back to the equation:

-(-12b-6)


We get rid of parentheses

-3b^2+3b+12b+6=0

We add all the numbers together, and all the variables

-3b^2+15b+6=0

a = -3; b = 15; c = +6;
Δ = b2-4ac
Δ = 152-4·(-3)·6
Δ = 297
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:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{33}}{2*-3}=\frac{-15-3\sqrt{33}}{-6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{33}}{2*-3}=\frac{-15+3\sqrt{33}}{-6}
$