2k^2+16k+24=0

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Solution for 2k^2+16k+24=0 equation: 



2k^2+16k+24=0

a = 2; b = 16; c = +24;
Δ = b2-4ac
Δ = 162-4·2·24
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*2}=\frac{-24}{4}
=-6
$


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

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