33j^2+7j=0

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Solution for 33j^2+7j=0 equation: 



33j^2+7j=0

a = 33; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·33·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*33}=\frac{-14}{66}
=-7/33
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*33}=\frac{0}{66}
=0
$

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