Minimize distance calculus

i keep getting an answer that has a second derivative that is negative, or concave down, meaning its a maximum not a minimum.
find the point on the line 4x + y = 6 that is closest to the point (-5, 3) to 2 decimal places.

4x + y = 6
y = 6 - 4x

distance to (-5,3):
sqrt( (x + 5)^2 + (6 - 4x - 3)^2)
sqrt( (x + 5)^2 + (3 - 4x)^2 )
sqrt( x^2 + 10x + 25 + 9 - 24x + 16x^2)
sqrt(17x^2 - 14x + 34)

first derivative:
(34x - 14) / 2 sqrt(17x^2 - 14x + 34)
(17x - 7) / sqrt(17x^2 - 14x + 34) = 0
17x - 7 = 0
x = 7/17 = .41
y = 6 - 4(.41) = 4.36

Taking the second derivative seems unnecessarily complicated; it seems easier to test a value on either side of x = .41 to confirm that it is a minimum. However, since your difficulty is with the second derivative, let's do it.

derivative of u'/2 * u^(-1/2) = u'/2 * (- u^(-3/2) u' / 2) + (u''/2)u^(-1/2)
(-(u')^2 u^(-3/2) + 2u'' u^(-1/2)) / 4
(2u'' u - (u')^2) / 4u^(3/2)

u = 17x^2 - 14x + 34
u' = 34x - 14
u'' = 34

(2(34)(17x^2 - 14x + 34) - (34x - 14)^2 )/ 4(17x^2 - 14x + 34)^(3/2)
(68(17x^2 - 14x + 34) - (34^2 x^2 - 2* 34 * 14 x + 14^2 x) ) / 4(17x^2 - 14x + 34)^(3/2)
(68(17x^2 - 14x + 34) - 68(17 x^2 - 14x + 49/17) ) / 4(17x^2 - 14x + 34)^(3/2)
(68(34 - 49/17) ) / 4(17x^2 - 14x + 34)^(3/2)
(17^2*8- 4*49) / 4(17x^2 - 14x + 34)^(3/2)
(2*17^2 - 49) / (17x^2 - 14x + 34)^(3/2)
529 / (17x^2 - 14x + 34)^(3/2)

529 / (17 * .41^2 - 14 * .41 + 32)^(3/2) = 3.37
second derivative is positive

Answer: (0.41, 4.35)

Not sure how you were going about doing this from the problem description, but here's what I would do:
In slope intercept form, y = -4x + 6

Set up an equation for the distance using the distance formula:

D = √( (x1 - x2)^2 + (y1 - y2)^2 )

Let the x coordinate of the closest point on the line be x. The y coordinate would then be -4x + 6