4k2+12k+8=0

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Solution for 4k2+12k+8=0 equation: 



4k^2+12k+8=0

a = 4; b = 12; c = +8;
Δ = b2-4ac
Δ = 122-4·4·8
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4}{2*4}=\frac{-16}{8}
=-2
$


$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*4}=\frac{-8}{8}
=-1
$

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