Math help? Quadratic Relations

zeros are -1 and 7, a point on the graph is 3,5
write the equation in standard form.
starting with y=a(x-s)(x-t)
I understand how to get a so a is -5/16 but i always simplify it wrong. Help?

(x+1)(x-7)=x^2-6x-7
y=a(x^2-6x-7)
5=a(3^2-6.3-7)
a=-5/16
y=-5/16x^2+15/8x+35/16
y=-5/16(x^2-6x +9-9) +35/16
y=-5/16(x-3)^2 +35/16+45/16
y=-5/16(x-3)^2 + 5

s and t are the zeroes, so
y = a(x+1)(x-7)
putting in x=3, y=5:
5 = a(4)(-4)
=> a = -5/16

EDIT: Sorry, you want standard form, i.e. ax^2 + bx + c. So you expand the brackets:

y = (-5/16)(x+1)(x-7)
= (-5/16)(x^2 + x - 7x - 7)
= (-5/16)(x^2 - 6x - 7)
= (-5/16)x^2 + (-5/16)(-6)x + (-5/16)(-7)
= (-5/16)x^2 + (30/16)x + 35/16
= (-5/16)x^2 + (15/8)x + 35/16

EDIT 2: Seems there are different interpretations of "standard form" out there, so I',m not sure if my answer is correct, or the other one