What is the center and radius for the circle with equation 2x^2-8x+2y^2+12y+14=0?

Complete the squaer for x's and y's seperately
actually, lets undsitribute the  2 for everybody first (except 14)

2(x^2-4x+y^2+6y)+14=0
group x and y's
2((x^2-4x)+(y^2+6y))+14=0
take 1/2 of the linear coefients and squaer them then add negative and positive of them inside
-4/2=-2, (-2)^2,  6/2=3, 3^2=9
2((x^2-4x+4-4)+(y^2+6y+9-9))+14=0
factor perfect squares
2(((x-2)^2-4)+((y+3)^2-9))+14=0
distribute
2(x-2)^2-8+2(y+3)^2-18+14=0
2(x-2)^2+2(y+3)^2-12=0
add 12 both sides
2(x-2)^2+2(y+3)^2=12
divide both sides by 2
(x-2)^2+(y+3)^2=6

for
(x-h)^2+(y-k)^2=r^2
center is (h,k)
radius is r

(x-2)^2+(y+3)^2=6
(x-2)^2+(y-(-3))^2=(√6)^2

center is (2,-3) and radius is√6