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- #1
- Feb 14, 2012
- 3,838
Solve the equation below for real solutions:
$\dfrac{1}{(x^2+x+2)^2}+\dfrac{1}{16}-\dfrac{1}{x^4-x^3+x^2+3x+4}=0$
$\dfrac{1}{(x^2+x+2)^2}+\dfrac{1}{16}-\dfrac{1}{x^4-x^3+x^2+3x+4}=0$
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