Solve Rational equations (Quadratic)

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Note that $x \neq4$ and $x \neq-1$

**Answer**

$\frac{127}{x-4}-\frac{2}{x+1}-\frac{5}{\left(x-4\right)\left(x+1\right)}=-20$

Or, $127\left(x+1\right)-2\left(x-4\right)-5=-20\left(x-4\right)\left(x+1\right)$

Or, $127x+127-2x+8-5=-20\left(x^{2}-3x-4\right)$

Or, $125x+130=-20x^{2}+60x+80$

Or, $125x+130+20x^{2}-60x-80=0$Or, $20x^{2}+65x+50=0$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*, $a=20$, $b=65$, and $c=50$

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

$x=\frac{-65\pm \sqrt{4225-4000}}{40}$

or, $x = \frac{-65\pm \sqrt{225}}{40}$

or, $x = \frac{-65\pm 15}{40}$

or, $x = \frac{-50}{40}$ and $x = \frac{-80}{40}$

or, $x=-\frac{5}{4}$ and $x =-2$

There are two distinct solutions given by $x = -\frac{5}{4}$ and $x = -2$