3k2+4k+1=0

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Solution for 3k2+4k+1=0 equation: 



3k^2+4k+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:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


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


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

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