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