Minimum value of equation

$\begin{aligned} P & = (x+3y-5)^2 - 6xy + 26 \\ & = x^2 + 9y^2 + 6xy - 10x - 30y + 25 - 6xy + 26 \\ & = x^2 + 9y^2 - 10x - 30y + 51 \\ & = (x^2 - 10x + 25) + (9y^2 - 30y + 25) + 1 \\ & = (x - 5)^2 + (3y - 5)^2 + 1 \end{aligned}$

Therefore,

$\begin{aligned} \min (P) & = \min \left((x - 5)^2 + (3y - 5)^2 + 1\right) \\ & = \min \left((x - 5)^2 \right) + \min \left((3y - 5)^2 \right) + 1 \\ & = 0 + 0 + 1 = \boxed 1 & \small \color{#3D99F6} \text{when }x=5, y = \frac 53 \end{aligned}$

There are no constraints, so partial differentiation is the easiest (but probably not the most elegant) way to solve this. We have

$P_x=2(x+3y-5)-6y=2(x-5)\\ P_y=6(x+3y-5)-6x=6(3y-5)$

Setting both of these equal to zero we get $x=5$ , $y=\frac53$ and substituting in, $P=\boxed1$ (it's easy to see from the form of $P$ that this must be a minimum).