Multiply both sides by \(x\).
Move all terms to one side.
Factor with quadratic formula.
In general, given \(a{x}^{2}+bx+c\), the factored form is:
\[a(x-\frac{-b+\sqrt{{b}^{2}-4ac}}{2a})(x-\frac{-b-\sqrt{{b}^{2}-4ac}}{2a})\]
In this case, \(a=1\), \(b=-2.5\) and \(c=1\).
\[(x-\frac{2.5+\sqrt{{(-2.5)}^{2}-4}}{2})(x-\frac{2.5-\sqrt{{(-2.5)}^{2}-4}}{2})\]
Simplify.
\[(x-2)(x-\frac{1}{2})\]
Solve for \(x\).
Ask: When will \((x-2)(x-\frac{1}{2})\) equal zero?
When \(x-2=0\) or \(x-\frac{1}{2}=0\)
Solve each of the 2 equations above.
\[x=2,\frac{1}{2}\]
