k2-10k+26=8

Solution for k2-10k+26=8 equation:

k2-10k+26=8

We move all terms to the left:

k2-10k+26-(8)=0

We add all the numbers together, and all the variables

k^2-10k+18=0

a = 1; b = -10; c = +18;
Δ = b2-4ac
Δ = -102-4·1·18
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{7}}{2*1}=\frac{10-2\sqrt{7}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{7}}{2*1}=\frac{10+2\sqrt{7}}{2}
$