Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 12:59Z
2ψ² + 5ψ + 4 = 0
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
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N₂ + 3H₂ ⇌ 2NH₃
\lim_{φ\to 0} \frac{\sin φ}{φ} = 1
∫₀^∞ e^(-ψ²) dψ = √π / 2
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∇²z = 0
N₂ + 3H₂ ⇌ 2NH₃
8t² + 2t + 3 = 0
det| 7 4 ; 4 8 | = 40
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