Solve Rational equations (Quadratic)

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Note that $x \neq4$ and $x \neq-5$

**Answer**

$\frac{95}{x-4}-\frac{257}{x+5}+\frac{27}{\left(x-4\right)\left(x+5\right)}=-81$

Or, $95\left(x+5\right)-257\left(x-4\right)+27=-81\left(x-4\right)\left(x+5\right)$

Or, $95x+475-257x+1028+27=-81\left(x^{2}+x-20\right)$

Or, $-162x+1530=-81x^{2}-81x+1620$

Or, $-162x+1530+81x^{2}+81x-1620=0$Or, $81x^{2}-81x-90=0$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*, $a=81$, $b=-81$, and $c=-90$

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

$x=\frac{81\pm \sqrt{6561+29160}}{162}$

or, $x = \frac{81\pm \sqrt{35721}}{162}$

or, $x = \frac{81\pm 189}{162}$

or, $x = \frac{270}{162}$ and $x = \frac{-108}{162}$

or, $x=\frac{5}{3}$ and $x =-\frac{2}{3}$

There are two distinct solutions given by $x = \frac{5}{3}$ and $x = -\frac{2}{3}$