21x^2+42x=0

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Solution for 21x^2+42x=0 equation:



21x^2+42x=0
a = 21; b = 42; c = 0;
Δ = b2-4ac
Δ = 422-4·21·0
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42}{2*21}=\frac{-84}{42} =-2 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42}{2*21}=\frac{0}{42} =0 $

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