b(11b+5)=0

Solution for b(11b+5)=0 equation:

b(11b+5)=0

We multiply parentheses

11b^2+5b=0

a = 11; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·11·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*11}=\frac{-10}{22}
=-5/11
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*11}=\frac{0}{22}
=0
$