17v^2+35v=0

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Solution for 17v^2+35v=0 equation:



17v^2+35v=0
a = 17; b = 35; c = 0;
Δ = b2-4ac
Δ = 352-4·17·0
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-35}{2*17}=\frac{-70}{34} =-2+1/17 $
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+35}{2*17}=\frac{0}{34} =0 $

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