41u^2+7u=0

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Solution for 41u^2+7u=0 equation: 



41u^2+7u=0

a = 41; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·41·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*41}=\frac{-14}{82}
=-7/41
$


$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*41}=\frac{0}{82}
=0
$

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