PH help please!?

HCOOH(aq) + H₂O(l) ⇌ HCOO⁻(aq) + H₃O⁺(aq) .... Ka = 1.80 × 10⁻⁴

At equilibrium :
Let [HCOO⁻] = [H₃O⁺] = y M
Then [HCOOH] = (0.110 - y) M

Ka = [HCOO⁻][H₃O⁺]/[HCOOH]
y²/(0.110 - y) = 1.80 × 10⁻⁴
y² + (1.80 × 10⁻⁴)y - (1.98 × 10⁻⁵) = 0
y = {-(1.80 × 10⁻⁴) ± √[(1.80 × 10⁻⁴) - 4(-1.98 × 10⁻⁵)]} / 2
y = 4.36 × 10⁻³ or y = -4.54 × 10⁻³ (rejected)

At equilibrium :
[H₃O⁺] = 4.36 × 10⁻³ M
pH = -log[H₃O⁺] = -log(4.36 × 10⁻³) = 2.36