(105-22k+k^2)-24=0

Solution for (105-22k+k^2)-24=0 equation:

(105-22k+k^2)-24=0

We get rid of parentheses

k^2-22k+105-24=0

We add all the numbers together, and all the variables

k^2-22k+81=0

a = 1; b = -22; c = +81;
Δ = b2-4ac
Δ = -222-4·1·81
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-4\sqrt{10}}{2*1}=\frac{22-4\sqrt{10}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+4\sqrt{10}}{2*1}=\frac{22+4\sqrt{10}}{2}
$