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I am trying to solve a quadratic equation very hard. Is there any other way to solve this without quadratic formula?

$$ x^2(-BE(F+C)^2(G+C)(A+C))))+x(C(F+C))\left [ EBA(D-H)-(G+C)(A+C)(B(D-H)+D(F+C)) \right ]+(ABC^2(D-H))=0 $$

A, B, C, D, E, F, G, H are just constants...i want to solve for x.

I tried quadratic formula but lost myself...

Is there any trick? Can someone help?

Thanks!

Patrick Stevens
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pipita
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  • Replace the coefficients with $K,L,M$, solve using quadratic formula and substitute back. I don't think there is a way which would be much easier. – Wojowu Jan 03 '16 at 16:29
  • It's hard to prove, but I'd assume that knowing two of the coefficients of the above would not yield the other; this would mean that you'd have to use the quadratic formula. It might simplify some, but probably not. – Milo Brandt Jan 03 '16 at 16:30
  • @Wojowu I had done it but didnt get anything...just a lot of calculations and then i got a really hard root – pipita Jan 03 '16 at 16:35
  • @MiloBrandt what do you mean with knowing 2 of the coefficients of the above would not yeld the other? – pipita Jan 03 '16 at 16:36
  • You could complete the square using substitutions for the coefficients, like Wojowu suggested. Ostensibly, though, you'd still be doing the quadratic formula. For equations that are as complicated as this, the quadratic formula is basically the only way to go. – zz20s Jan 03 '16 at 16:36
  • @pipita I mean that you could presumably choose A, B, C, D, E, F, G and H such that the quadratic had the form $ax^2+bx+c$ for any given a, b, and c. That is, your quadratic represents a general quadratic, which makes the prospects for simplification less hopeful. (However, if you just plug everything into the quadratic formula, it is possible that you'd find a simplification) – Milo Brandt Jan 03 '16 at 16:46

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