A student mistakenly calculated pH of a 1.0 x 10^-7 M HI solution to be 7.0. Explain why student is wrong and calculate correct pH.?

Solution by using ICE table:

Before autoionization of water,
initial concentration of H₃O⁺, [H₃O⁺]ₒ = 1.0 × 10⁻⁷ M

_                 2H₂O(ℓ)  ⇌  H₃O⁺(aq)  +  OH⁻(aq)  Kw = 1.0 × 10⁻¹⁴
Initial (M):       --           1.0 × 10⁻⁷          0
Change (M):   --              +y                 +y
Eqm (M):        --      (1.0 × 10⁻⁷ + y)       y

At equilibrium :
Kw = [H₃O⁺] [OH⁻]
1.0 × 10⁻¹⁴ = (1.0 × 10⁻⁷ + y) y
y² + (1.0 × 10⁻⁷)y - (1.0 × 10⁻¹⁴) = 0
y = {-(1.0 × 10⁻⁷) + √[(1.0 × 10⁻⁷) + 4×(1.0 × 10⁻¹⁴)]} / 2
y = 6.18 × 10⁻⁸

pH at equilibrium = -log[H₃O⁺] = -log[(1.0 × 10⁻⁷) + (6.18 × 10⁻⁸)] = 6.8

I discovered how to get answer:
[H3O][OH]=10^-14

[x+10^-7][x]=10^-14
x=6.18E-8
H3O= 6.18E-8+ 10^-7

pH=6.79

I was trying to solve it with ICE method but didn't work... I guess I needed to think outside the box. Unfortunately I had to look it up: H+ of extremely dilute strong acids.