The quadratic formula is neat! The (informal) proof is very neat also. If you remember back to completing the square...
Tadaaaa!
As an alternative, you can multiply through by 4a.
ax^2 + bx +c =0 (a≠0)• 4a^2 x^2 +4abx +4ac =0
• 4a^2 x^2 +4abx = (2ax+b)^2 - b^2
• (2ax+b)^2 - b^2 +4ac =0
• (2ax+b)^2 = b^2 -4ac
• 2ax+b = ±√(b^2-4ac)
Therefore:
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Imo, this version is slightly less messy.