k^2+23=12k

Solution for k^2+23=12k equation:

k^2+23=12k

We move all terms to the left:

k^2+23-(12k)=0

a = 1; b = -12; c = +23;
Δ = b2-4ac
Δ = -122-4·1·23
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{13}}{2*1}=\frac{12-2\sqrt{13}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{13}}{2*1}=\frac{12+2\sqrt{13}}{2}
$