$$5x^2+28x = 49x$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 21 }{ 5 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 5x^2+28x &= 49x&& \text{move all terms to the left hand side } \\[1 em]5x^2+28x-49x &= 0&& \text{simplify left side} \\[1 em]5x^2-21x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 5x^{2}-21x = 0 } $, first we need to factor our $ x $.
$$ 5x^{2}-21x = x \left( 5x-21 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 5x-21 = 0$.
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