3a2+4a+1=0

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Solution for 3a2+4a+1=0 equation: 



3a^2+4a+1=0

a = 3; b = 4; c = +1;
Δ = b2-4ac
Δ = 42-4·3·1
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*3}=\frac{-6}{6}
=-1
$


$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*3}=\frac{-2}{6}
=-1/3
$

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