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