3x^2+27x+1=0

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Solution for 3x^2+27x+1=0 equation:



3x^2+27x+1=0
a = 3; b = 27; c = +1;
Δ = b2-4ac
Δ = 272-4·3·1
Δ = 717
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-\sqrt{717}}{2*3}=\frac{-27-\sqrt{717}}{6} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{717}}{2*3}=\frac{-27+\sqrt{717}}{6} $

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