21b+63b^2=0

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Solution for 21b+63b^2=0 equation:



21b+63b^2=0
a = 63; b = 21; c = 0;
Δ = b2-4ac
Δ = 212-4·63·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-21}{2*63}=\frac{-42}{126} =-1/3 $
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+21}{2*63}=\frac{0}{126} =0 $

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