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Question Number 172611 by Mikenice last updated on 29/Jun/22

Answered by MJS_new last updated on 29/Jun/22

$$\mathrm{I}\mathrm{can}\mathrm{solve}\mathrm{this}\mathrm{but}\mathrm{I}\mathrm{guess}{\mathrm{ir}}^{\prime }\mathrm{s}\mathrm{meant}\mathrm{to}\mathrm{be}$$$$\frac{x^2+4x}{3}+\frac{84}{x^2+4x}=11$$$$t=x^2+4x$$$$\frac{t}{3}+\frac{84}{t}=11$$$$t^2-33t+252=0$$$$t=12\vee t=21$$$$x=-7\vee x=-6\vee x=2\vee x=3$$

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