3v^2+17v+19=0

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Solution for 3v^2+17v+19=0 equation: 



3v^2+17v+19=0

a = 3; b = 17; c = +19;
Δ = b2-4ac
Δ = 172-4·3·19
Δ = 61
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-\sqrt{61}}{2*3}=\frac{-17-\sqrt{61}}{6}
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{61}}{2*3}=\frac{-17+\sqrt{61}}{6}
$

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