[(x^2)-20x+(23/3)]'-√576+4!=0

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[(x^2)-20x+(23/3)]'-√576+4!=0

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what is the answer to [(x^2)-20x+(23/3)]'-√576+4!=0?

Guest Aug 19, 2015

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[(x^2)-20x+(23/3)]'-√576+4!=0

$\\\left(x^2-20x+\frac{23}{3}\right)-\sqrt{576}+4!=0\\\\ \left(x^2-20x+\frac{23}{3}\right)-24+24=0\\\\ x^2-20x+\frac{23}{3}=0\\\\ 3x^2-60x+23=0\\\\ x=\frac{60\pm\sqrt{3600-276}}{6}\\\\ x=\frac{60\pm\sqrt{3324}}{6}\\\\ etc$

Melody Aug 19, 2015

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Melody · Answer 1 · 2015-08-19T01:17:09

[(x^2)-20x+(23/3)]'-√576+4!=0

$\\\left(x^2-20x+\frac{23}{3}\right)-\sqrt{576}+4!=0\\\\ \left(x^2-20x+\frac{23}{3}\right)-24+24=0\\\\ x^2-20x+\frac{23}{3}=0\\\\ 3x^2-60x+23=0\\\\ x=\frac{60\pm\sqrt{3600-276}}{6}\\\\ x=\frac{60\pm\sqrt{3324}}{6}\\\\ etc$

Melody · Answer 2 · 2015-08-19T12:55:42

[(x^2)-20x+(23/3)]'-√576+4!=0?

What is the apostrophy for?

Guest · Answer 3 · 2015-08-19T01:09:50

sorry must have put that there by accident...

[(x^2)-20x+(23/3)]'-√576+4!=0

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