Solve Rational equations (Quadratic)

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

**Answer**

$\frac{155}{x-3}-\frac{197}{x+3}+\frac{6}{\left(x-3\right)\left(x+3\right)}=-120$

Or, $155\left(x+3\right)-197\left(x-3\right)+6=-120\left(x-3\right)\left(x+3\right)$

Or, $155x+465-197x+591+6=-120\left(x^{2}-9\right)$

Or, $-42x+1062=-120x^{2}+1080$

Or, $-42x+1062+120x^{2}-1080=0$Or, $120x^{2}-42x-18=0$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*, $a=120$, $b=-42$, and $c=-18$

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

$x=\frac{42\pm \sqrt{1764+8640}}{240}$

or, $x = \frac{42\pm \sqrt{10404}}{240}$

or, $x = \frac{42\pm 102}{240}$

or, $x = \frac{144}{240}$ and $x = \frac{-60}{240}$

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

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