19y2+22y=0

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Solution for 19y2+22y=0 equation:



19y^2+22y=0
a = 19; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·19·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*19}=\frac{-44}{38} =-1+3/19 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*19}=\frac{0}{38} =0 $

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