Given: #4x^2-5y^2-40x-20y+160=0#

Subtract 160 from both sides:

#4x^2-5y^2-40x-20y=-160#

Add #4h^2# to both sides:

#4x^2-40x+4h^2-5y^2-20y=-160+4h^2#

Remove the common factor, 4, from the first 3 terms:

#4(x^2-10x+h^2)-5y^2-20y=-160+4h^2#

Subtract #5k^2# from both sides:

#4(x^2-10x+h^2)-5y^2-20y-5k^2=-160+4h^2-5k^2#

Remove a common factor of -5 from the next 3 terms:

#4(x^2-10x+h^2)-5(y^2+4y+k^2)=-160+4h^2-5k^2" [1]"#

Because we are looking for the square #(x-h)^2 = x^2-2hx+h^2# we use the equation:

#-2hx = -10x#

To find the value of h:

#h = 5#

Into equation [1], substitute #(x-5)^2" for " x^2-10x+h^2# and 100 for #4h^2#:

#4(x-5)^2-5(y^2+4y+k^2)=-160+100-5k^2" [1.1]"#

Because we are looking for the square #(y-k)^2 = y^2-2ky+k^2# we use the equation:

#-2ky = 4y#

To find the value of k:

#k = -2#

Into equation [1.1], substitute #(y-(-2))^2" for " y^2+4y+k^2# and -20 for #-5k^2#:

#4(x-5)^2-5(y-(-2))^2=-160+100-20" [1.2]"#

Combine like terms on the right:

#4(x-5)^2-5(y-(-2))^2=-80" [1.3]"#

Divide both sides by -80:

#(y-(-2))^2/16-(x-5)^2/20=1" [1.3]"#

Write the denominators as squares:

#(y-(-2))^2/(4)^2-(x-5)^2/(2sqrt5)^2=1" [1.4]"#

Equation [1.4] is standard form.