4q^2+4q=0

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Solution for 4q^2+4q=0 equation: 



4q^2+4q=0

a = 4; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·4·0
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*4}=\frac{-8}{8}
=-1
$


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

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