21k^2+7k=0

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Solution for 21k^2+7k=0 equation: 



21k^2+7k=0

a = 21; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·21·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*21}=\frac{-14}{42}
=-1/3
$


$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*21}=\frac{0}{42}
=0
$

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