-108=-b(1+3b)

Solution for -108=-b(1+3b) equation:

-108=-b(1+3b)

We move all terms to the left:

-108-(-b(1+3b))=0

We add all the numbers together, and all the variables

-(-b(3b+1))-108=0


We calculate terms in parentheses: -(-b(3b+1)), so:

-b(3b+1)

We multiply parentheses

-3b^2-b

We add all the numbers together, and all the variables

-3b^2-1b

Back to the equation:

-(-3b^2-1b)


We get rid of parentheses

3b^2+1b-108=0

We add all the numbers together, and all the variables

3b^2+b-108=0

a = 3; b = 1; c = -108;
Δ = b2-4ac
Δ = 12-4·3·(-108)
Δ = 1297
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{1297}}{2*3}=\frac{-1-\sqrt{1297}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1297}}{2*3}=\frac{-1+\sqrt{1297}}{6}
$