v^2+22v=0

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Solution for v^2+22v=0 equation: 



v^2+22v=0

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

$\sqrt{\Delta}=\sqrt{484}=22$


$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*1}=\frac{-44}{2}
=-22
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*1}=\frac{0}{2}
=0
$

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