b(11+6b)=35

Solution for b(11+6b)=35 equation:

b(11+6b)=35

We move all terms to the left:

b(11+6b)-(35)=0

We add all the numbers together, and all the variables

b(6b+11)-35=0

We multiply parentheses

6b^2+11b-35=0

a = 6; b = 11; c = -35;
Δ = b2-4ac
Δ = 112-4·6·(-35)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-31}{2*6}=\frac{-42}{12}
=-3+1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+31}{2*6}=\frac{20}{12}
=1+2/3
$