6x2+19x+3=0

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Solution for 6x2+19x+3=0 equation:



6x^2+19x+3=0
a = 6; b = 19; c = +3;
Δ = b2-4ac
Δ = 192-4·6·3
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-17}{2*6}=\frac{-36}{12} =-3 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+17}{2*6}=\frac{-2}{12} =-1/6 $

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