$$\frac{1+x}{2}+\frac{3-x}{4} = x^2+4$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 8 }+\dfrac{ 5 \sqrt{ 7}}{ 8 }i & x_2 = \dfrac{ 1 }{ 8 }-\dfrac{ 5 \sqrt{ 7}}{ 8 }i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1+x}{2}+\frac{3-x}{4} &= x^2+4&& \text{multiply ALL terms by } \color{blue}{ 4 }. \\[1 em]4 \cdot \frac{1+x}{2}+4\frac{3-x}{4} &= 4x^2+4\cdot4&& \text{cancel out the denominators} \\[1 em]2x+2+3-x &= 4x^2+16&& \text{simplify left side} \\[1 em]x+5 &= 4x^2+16&& \text{move all terms to the left hand side } \\[1 em]x+5-4x^2-16 &= 0&& \text{simplify left side} \\[1 em]-4x^2+x-11 &= 0&& \\[1 em] \end{aligned} $$
$ -4x^{2}+x-11 = 0 $ is a quadratic equation.
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