Question
$$\frac{1+x}{2}+\frac{3-x}{4} = (x-3)^2+2$$
Answer
$$ \begin{matrix}x_1 = 3 & x_2 = \dfrac{ 13 }{ 4 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1+x}{2}+\frac{3-x}{4} &= (x-3)^2+2&& \text{multiply ALL terms by } \color{blue}{ 4 }. \\[1 em]4 \cdot \frac{1+x}{2}+4\frac{3-x}{4} &= 4(x-3)^2+4\cdot2&& \text{cancel out the denominators} \\[1 em]2x+2+3-x &= 4(x-3)^2+8&& \text{simplify left and right hand side} \\[1 em]x+5 &= 4(x^2-6x+9)+8&& \\[1 em]x+5 &= 4x^2-24x+36+8&& \\[1 em]x+5 &= 4x^2-24x+44&& \text{move all terms to the left hand side } \\[1 em]x+5-4x^2+24x-44 &= 0&& \text{simplify left side} \\[1 em]-4x^2+25x-39 &= 0&& \\[1 em] \end{aligned} $$
$ -4x^{2}+25x-39 = 0 $ is a quadratic equation.
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