6q^2+6q=0

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Solution for 6q^2+6q=0 equation: 



6q^2+6q=0

a = 6; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·6·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*6}=\frac{-12}{12}
=-1
$


$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*6}=\frac{0}{12}
=0
$

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