Solve Rational equations (Quadratic)

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Note that $x \neq3$ and $x \neq-3$

**Answer**

$\frac{119}{x-3}-\frac{68}{x+3}-\frac{12}{\left(x-3\right)\left(x+3\right)}=-60$

Or, $119\left(x+3\right)-68\left(x-3\right)-12=-60\left(x-3\right)\left(x+3\right)$

Or, $119x+357-68x+204-12=-60\left(x^{2}-9\right)$

Or, $51x+549=-60x^{2}+540$

Or, $51x+549+60x^{2}-540=0$Or, $60x^{2}+51x+9=0$The solutions of a *quadratic equation* $ax^2+bx + c = 0$ are given by$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

In this *quadratic equation*, $a=60$, $b=51$, and $c=9$

Substituting these values of $a$, $b$ and $c$ in the above formula, we get

$x=\frac{-51\pm \sqrt{2601-2160}}{120}$

or, $x = \frac{-51\pm \sqrt{441}}{120}$

or, $x = \frac{-51\pm 21}{120}$

or, $x = \frac{-30}{120}$ and $x = \frac{-72}{120}$

or, $x=-\frac{1}{4}$ and $x =-\frac{3}{5}$

There are two distinct solutions given by $x = -\frac{1}{4}$ and $x = -\frac{3}{5}$