Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:51Z

11n² + 5n + 4 = 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

460K

e^{i\pi} + 1 = 0

438K

∫₀^∞ e^(-x²) dx = √π / 2

6ω² + 4ω + 4 = 0

186K

12y² + 5y + 7 = 0

411K

4φ² + 6φ + 2 = 0

det| 7 4 ; 4 8 | = 40

448K

\lim_{α\to 0} \frac{\sin α}{α} = 1

22K

∑_{k=1}^{n} k = n(n+1)/2

459K

iħ ∂ψ/∂t = Ĥψ

406K

5φ² + 1φ + 1 = 0

181K

N₂ + 3H₂ ⇌ 2NH₃

80K

10ω² + 7ω + 4 = 0

359K