2y^2-17y+31=(y-7)^2 solve for y?

2y² - 17y + 31 = (y - 7)²........
=> 2y² - 17y + 31 = y² + 49 - 14y
=> 2y² - y² = 49 - 31 + 17y - 14y
=> y² = 18 + 3y
=> y² - 3y - 18 = 0
=> y² - 6y + 3y - 18 = 0
=> y(y - 6)+ 3(y - 6) = 0
=> y + 3 = 0 or y - 6 = 0
So, y = -3, 6. Hope that helped........

2y² - 17y + 31 = y² - 14y + 49
y² - 3y - 18 = 0
(y - 6)(y + 3) = 0
y = 6 , y = - 3

2y^2 - 17y + 31 = (y - 7)^2
2y^2 - 17y + 31 = (y - 7)(y - 7)
2y^2 - 17y + 31 = y*y - 7*y - y*7 + 7*7
2y^2 - 17y + 31 = y^2 - 7y - 7y + 49
2y^2 - 17y + 31 = y^2 - 14y + 49
2y^2 - y^2 - 17y + 14y + 31 - 49 = 0
y^2 - 3y - 18 = 0
y^2 + 3y - 6y - 18 = 0
(y^2 + 3y) - (6y + 18) = 0
y(y + 3) - 6(y + 3) = 0
(y + 3)(y - 6) = 0

y + 3 = 0
y = -3

y - 6 = 0
y = 6

∴ y = -3 , 6

Assuming that this is a homework question you should probably figure it out yourself, but:

Firstly expand (y-7)^2 to produce:
2y^2-17y+31=y^2-14y+49

Simplify to the form Ay^2+By+C=0:
y^2-3y-18=0

Now either factorise the equation or if you are smart, use the quadratic formula to find the roots of the equation and thus solve for y (note that +/- is supposed to represent the plus-minus symbol - you have to calculate it twice):

using y=(-b(+/-)(b^2-4ac)^1/2)/(2a) -->

(3+/-(9-4(-18))^1/2)/2 => (3+/-(81)^1/2)/2 => (3+/-9)/2

thus y = 6 and -3, sub into original equation to confirm.

Remember that the quadratic formula will return imaginary numbers (i = -1^1/2) if b^2-4ac returns a negative number. This means either you've made a mistake, or the equation cannot be factorised in x-y co-ordinate space.

2y^2-17y+31=(y-7)^2
2y^2-17y+31=y^2-14y+49
0=-y^2+3y+18
0=y^2-3y-18
0=(y+3)(y-6)

0=y+3
y=-3

0=y-6
y=6

2y^2-17y+31-y^2-49+14y=0, y^2-3y-18=0,(y-9)(y+6)=0
y=9 & y=-6

2y^2-17y+31=y^2-14y+49
y^2-3y-18=0
(y+3)(y-6)=0
y=-3 or y = 6