Find the minmum value of x^2-4x+y^2+8y+20=?
we don’t know what it’s equal to, so we set it equal to ?, if you know that ? is, then input it
what we can do is try to convert it to a conic section
complete the squares

(x^2-4x)+(y^2+8y)+20=?
take 1/2 of linear coefient and square it and add negative and positive insde (linear is 1st degree)

(x^2-4x+4-4)+(y^2+8x+16-16)+20=?
factor perfect squares
((x-2)^2-4)+((y+4)^2-16)+20=?
(x-2)^2-4+(y+4)^2-16+20=?
(x-2)^2+(y+4)^2=?

we see this is the equation of a circle
in form
(x-h)^2+(y-k)^2=r^2
center is (h,k)
radius is r
so the lowest point is r  units down from (h,k), or the point (h,k-r)

we know that te equation is
(x-2)^2+(y-(-4))^2=?
?=r^2
√?=r
center is (2,-4)
therefor
the minimum value, where the equation is equal to ?, is (2,-4-√?)
good luck, and may the force be with you