$$5(s+7)^2-14(s+7) = 3$$
Answer
$$ \begin{matrix}s_1 = -4 & s_2 = -\dfrac{ 36 }{ 5 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 5(s+7)^2-14(s+7) &= 3&& \text{simplify left side} \\[1 em]5(1s^2+14s+49)-14(s+7) &= 3&& \\[1 em]5s^2+70s+245-(14s+98) &= 3&& \\[1 em]5s^2+70s+245-14s-98 &= 3&& \\[1 em]5s^2+56s+147 &= 3&& \text{move all terms to the left hand side } \\[1 em]5s^2+56s+147-3 &= 0&& \text{simplify left side} \\[1 em]5s^2+56s+144 &= 0&& \\[1 em] \end{aligned} $$
$ 5x^{2}+56x+144 = 0 $ is a quadratic equation.
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