$$4-x = \frac{15}{4x}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 16 }{ 19 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 4-x &= \frac{15}{4x}&& \text{multiply ALL terms by } \color{blue}{ 4x }. \\[1 em]4x\cdot4-4xx &= 4x\cdot\frac{15}{4x}&& \text{cancel out the denominators} \\[1 em]16x-4x^2 &= 15x^2&& \text{simplify left side} \\[1 em]-4x^2+16x &= 15x^2&& \text{move all terms to the left hand side } \\[1 em]-4x^2+16x-15x^2 &= 0&& \text{simplify left side} \\[1 em]-19x^2+16x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -19x^{2}+16x = 0 } $, first we need to factor our $ x $.
$$ -19x^{2}+16x = x \left( -19x+16 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -19x+16 = 0$.
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