$$\frac{49}{10}x^2+19x = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 190 }{ 49 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{49}{10}x^2+19x &= 0&& \text{multiply ALL terms by } \color{blue}{ 10 }. \\[1 em]10 \cdot \frac{49}{10}x^2+10\cdot19x &= 10\cdot0&& \text{cancel out the denominators} \\[1 em]49x^2+190x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 49x^{2}+190x = 0 } $, first we need to factor our $ x $.
$$ 49x^{2}+190x = x \left( 49x+190 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 49x+190 = 0$.
This page was created using
Polynomial Equations Solver