$$\frac{k+4}{4}+\frac{k-1}{4} = \frac{k+4}{4k}$$
Answer
$$ \begin{matrix}k_1 = 0 & k_2 = 1 & k_3 = -3 \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{k+4}{4}+\frac{k-1}{4} &= \frac{k+4}{4k}&& \text{multiply ALL terms by } \color{blue}{ 4k }. \\[1 em]4k\frac{k+4}{4}+4k\frac{k-1}{4} &= 4k\frac{k+4}{4k}&& \text{cancel out the denominators} \\[1 em]k^2+4k+k^2-k &= k^3+4k^2&& \text{simplify left side} \\[1 em]2k^2+3k &= k^3+4k^2&& \text{move all terms to the left hand side } \\[1 em]2k^2+3k-k^3-4k^2 &= 0&& \text{simplify left side} \\[1 em]-k^3-2k^2+3k &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -x^{3}-2x^{2}+3x = 0 } $, first we need to factor our $ x $.
$$ -x^{3}-2x^{2}+3x = x \left( -x^{2}-2x+3 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ -x^{2}-2x+3 = 0$.
$ -x^{2}-2x+3 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
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