Field Notes / Σ /sym/b26508
id=1 · host=uk.mattivster.me · 2026-10-19 23:28Z
3θ² + 8θ + 7 = 0
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
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∇ × E = −∂B/∂t
\frac{12}{3} \sum_{i=1}^{n} i^{2}
λ = h/p
\binom{n}{k} = \frac{n!}{k!(n-k)!}
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det| 9 4 ; 1 10 | = 86
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iħ ∂ψ/∂t = Ĥψ
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
d/dβ [β^11] = 11β^10