$$\frac{6x^2+22x-5}{x+4} = 0$$
Answer
$$ \begin{matrix}x_1 = -\dfrac{ 11 }{ 6 }-\dfrac{\sqrt{ 151 }}{ 6 } & x_2 = -\dfrac{ 11 }{ 6 }+\dfrac{\sqrt{ 151 }}{ 6 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{6x^2+22x-5}{x+4} &= 0&& \text{multiply ALL terms by } \color{blue}{ x+4 }. \\[1 em](x+4)\frac{6x^2+22x-5}{x+4} &= (x+4)\cdot0&& \text{cancel out the denominators} \\[1 em]6x^2+22x-5 &= 0&& \\[1 em] \end{aligned} $$
$ 6x^{2}+22x-5 = 0 $ is a quadratic equation.
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