3k^2+12k+9=0

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Solution for 3k^2+12k+9=0 equation: 



3k^2+12k+9=0

a = 3; b = 12; c = +9;
Δ = b2-4ac
Δ = 122-4·3·9
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6}{2*3}=\frac{-18}{6}
=-3
$


$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6}{2*3}=\frac{-6}{6}
=-1
$

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