k2+28=16k

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Solution for k2+28=16k equation:



k2+28=16k
We move all terms to the left:
k2+28-(16k)=0
We add all the numbers together, and all the variables
k^2-16k+28=0
a = 1; b = -16; c = +28;
Δ = b2-4ac
Δ = -162-4·1·28
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-12}{2*1}=\frac{4}{2} =2 $
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+12}{2*1}=\frac{28}{2} =14 $

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