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If {a+b=40b+c=3a+c=−11 \begin{cases} a + b = 40 \\ b + c = 3 \\ a + c = -11 \end{cases} ⎩⎪⎨⎪⎧a+b=40b+c=3a+c=−11 where aaa, bbb and ccc are real numbers, and that a+b+c=k a + b + c = k a+b+c=k, then find xxx that satisfies the following equation.
kx24−44kx+1933k=−48 \frac{kx^2}{4} - 44kx + 1933k = -48 4kx2−44kx+1933k=−48
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