Question
$$(7y+2)^2+112 = 0$$
Answer
$$ \begin{matrix}y_1 = -\dfrac{ 2 }{ 7 }+\dfrac{ 4 \sqrt{ 7}}{ 7 }i & y_2 = -\dfrac{ 2 }{ 7 }-\dfrac{ 4 \sqrt{ 7}}{ 7 }i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (7y+2)^2+112 &= 0&& \text{simplify left side} \\[1 em]49y^2+28y+4+112 &= 0&& \\[1 em]49y^2+28y+116 &= 0&& \\[1 em] \end{aligned} $$
$ 49x^{2}+28x+116 = 0 $ is a quadratic equation.
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