(6q-1)(q+11)=0

Solution for (6q-1)(q+11)=0 equation:

(6q-1)(q+11)=0

We multiply parentheses ..

(+6q^2+66q-1q-11)=0

We get rid of parentheses

6q^2+66q-1q-11=0

We add all the numbers together, and all the variables

6q^2+65q-11=0

a = 6; b = 65; c = -11;
Δ = b2-4ac
Δ = 652-4·6·(-11)
Δ = 4489
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4489}=67$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(65)-67}{2*6}=\frac{-132}{12}
=-11
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(65)+67}{2*6}=\frac{2}{12}
=1/6
$