$$\frac{42x^2-33}{7x+7} = 0$$
Answer
$$ \begin{matrix}x_1 = - \dfrac{\sqrt{ 154 }}{ 14 } & x_2 = \dfrac{\sqrt{ 154 }}{ 14 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{42x^2-33}{7x+7} &= 0&& \text{multiply ALL terms by } \color{blue}{ 7x+7 }. \\[1 em](7x+7)\frac{42x^2-33}{7x+7} &= (7x+7)\cdot0&& \text{cancel out the denominators} \\[1 em]42x^2-33 &= 0&& \\[1 em] \end{aligned} $$
$ 42x^{2}-33 = 0 $ is a quadratic equation.
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