3n2+11n+4=0

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Solution for 3n2+11n+4=0 equation: 



3n^2+11n+4=0

a = 3; b = 11; c = +4;
Δ = b2-4ac
Δ = 112-4·3·4
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{73}}{2*3}=\frac{-11-\sqrt{73}}{6}
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{73}}{2*3}=\frac{-11+\sqrt{73}}{6}
$

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