Here's the question you clicked on:
itsladyrw
Solve for x: ((2 / x - 2) + ( 7x^2 -4)) = 5/x
\[((\frac{ 2 }{ x }-2) + (7x ^{2}-4)) = \frac{ 5 }{ x }\] so this is your initial equation. as there aren't any multipliers outside the parenthesis, they can actually be left out: \[\frac{ 2 }{ x }-2 + 7x ^{2}-4 = \frac{ 5 }{ x }\] then, combine like terms: \[\frac{ 2 }{ x } + 7x ^{2}-6 = \frac{ 5 }{ x }\] in this case, i would multiply both sides by X to eliminate the fractions: \[2+ 7x^{3}-6x=5\] subtract 5 from both sides: \[ 7x^{3}-6x - 3=0\] then you can graph to find the zeros. is that what you're looking for?