1 F.4 Maths Question ONLY , Thanks !!!

Suppose the equation x^4-8^3+16x^2-4=0 can be transformed to the form y^4-8y^2+k=0 by substituting x=y+h
(a) Find the values of h and k
(b) Hence solve the equation x^4-8x^3+16x^2-4=0
With steps, Thanks
(a)
sub x=y+h, into x^4-8x^3+16x^2-4=0
(y+h)^4-8(y+h)^3+16(y+h)^2-4=0
y^4+4y^3h+6y^2h^2+4yh^3+h^4
-8(y^3+3y^2h+3yh^2+h^3)+16(y^2+2yh+h^2)-4=0
y^4+4y^3h+6y^2h^2+4yh^3+h^4
+(-8y^3-24y^2h-24yh^2-8h^3)+(16y^2+32yh+16h^2)-4=0
the coefficient of y^3 is 4h-8
so h=2
y^4+4y^3h+6y^2h^2+4yh^3+h^4
+(-8y^3-24y^2h-24yh^2-8h^3)+(16y^2+32yh+16h^2)-4=0
becomes
y^4+8y^3+24y^2+32y+16
+(-8y^3-48y^2-96y-64)+(16y^2+64h+64)-4=0

y^4-8y^2+12=0
k=12
(b)
x^4-8x^3+16x^2-4=0

y^4-8y^2+12=0
(y^2-6)(y^2-2)=0
y^2=2 or 6
y=√2 or -√2 or √6 or -√6
x=y+2
So
x=√2+2 or 2 -√2 or √6+2 or 2 -√6