(q)(11q+4)=0

Solution for (q)(11q+4)=0 equation:

(q)(11q+4)=0

We multiply parentheses

11q^2+4q=0

a = 11; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·11·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*11}=\frac{-8}{22}
=-4/11
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*11}=\frac{0}{22}
=0
$