Solve Rational equations (Quadratic)

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Note that $x \neq1$ and $x \neq3$

**Answer**

$\frac{15}{x-1}+\frac{89}{x-3}+\frac{12}{\left(x-1\right)\left(x-3\right)}=-30$

Or, $15\left(x-3\right)+89\left(x-1\right)+12=-30\left(x-1\right)\left(x-3\right)$

Or, $15x-45+89x-89+12=-30\left(x^{2}-4x+3\right)$

Or, $104x-122=-30x^{2}+120x-90$

Or, $104x-122+30x^{2}-120x+90=0$Or, $30x^{2}-16x-32=0$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*, $a=30$, $b=-16$, and $c=-32$

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

$x=\frac{16\pm \sqrt{256+3840}}{60}$

or, $x = \frac{16\pm \sqrt{4096}}{60}$

or, $x = \frac{16\pm 64}{60}$

or, $x = \frac{80}{60}$ and $x = \frac{-48}{60}$

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

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