$$4x^3+3x^4-(x^4-5x^3) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 9 }{ 2 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 4x^3+3x^4-(x^4-5x^3) &= 0&& \text{simplify left side} \\[1 em]4x^3+3x^4-x^4+5x^3 &= 0&& \\[1 em]2x^4+9x^3 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 2x^{4}+9x^{3} = 0 } $, first we need to factor our $ x^3 $.
$$ 2x^{4}+9x^{3} = x^3 \left( 2x+9 \right) $$$ x = 0 $ is a root of multiplicity $ 3 $.
The second root can be found by solving equation $ 2x+9 = 0$.
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