b+11=3b(b-1)

Solution for b+11=3b(b-1) equation:

b+11=3b(b-1)

We move all terms to the left:

b+11-(3b(b-1))=0


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

3b(b-1)

We multiply parentheses

3b^2-3b

Back to the equation:

-(3b^2-3b)


We get rid of parentheses

-3b^2+b+3b+11=0

We add all the numbers together, and all the variables

-3b^2+4b+11=0

a = -3; b = 4; c = +11;
Δ = b2-4ac
Δ = 42-4·(-3)·11
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{37}}{2*-3}=\frac{-4-2\sqrt{37}}{-6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{37}}{2*-3}=\frac{-4+2\sqrt{37}}{-6}
$