3q^2+7q=0

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Solution for 3q^2+7q=0 equation: 



3q^2+7q=0

a = 3; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·3·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*3}=\frac{-14}{6}
=-2+1/3
$


$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*3}=\frac{0}{6}
=0
$

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