3x2+19x+16=0

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Solution for 3x2+19x+16=0 equation:



3x^2+19x+16=0
a = 3; b = 19; c = +16;
Δ = b2-4ac
Δ = 192-4·3·16
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*3}=\frac{-32}{6} =-5+1/3 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*3}=\frac{-6}{6} =-1 $

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