k2+28=16k

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
$