## itsladyrw 3 years ago 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?