-2v^2+3v+2=0

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Solution for -2v^2+3v+2=0 equation: 



-2v^2+3v+2=0

a = -2; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·(-2)·2
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*-2}=\frac{-8}{-4}
=+2
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*-2}=\frac{2}{-4}
=-1/2
$

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