k^2-22k+7=4

Solution for k^2-22k+7=4 equation:

k^2-22k+7=4

We move all terms to the left:

k^2-22k+7-(4)=0

We add all the numbers together, and all the variables

k^2-22k+3=0

a = 1; b = -22; c = +3;
Δ = b2-4ac
Δ = -222-4·1·3
Δ = 472
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:

$\sqrt{\Delta}=\sqrt{472}=\sqrt{4*118}=\sqrt{4}*\sqrt{118}=2\sqrt{118}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{118}}{2*1}=\frac{22-2\sqrt{118}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{118}}{2*1}=\frac{22+2\sqrt{118}}{2}
$