$$\frac{1+x}{2}+\frac{3-x}{4} = 2(x+2)\cdot2(x-3)$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 17 }{ 32 }-\dfrac{\sqrt{ 6753 }}{ 32 } & x_2 = \dfrac{ 17 }{ 32 }+\dfrac{\sqrt{ 6753 }}{ 32 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1+x}{2}+\frac{3-x}{4} &= 2(x+2)\cdot2(x-3)&& \text{multiply ALL terms by } \color{blue}{ 4 }. \\[1 em]4 \cdot \frac{1+x}{2}+4\frac{3-x}{4} &= 42(x+2)\cdot2(x-3)&& \text{cancel out the denominators} \\[1 em]2x+2+3-x &= 16x^2-16x-96&& \text{simplify left side} \\[1 em]x+5 &= 16x^2-16x-96&& \text{move all terms to the left hand side } \\[1 em]x+5-16x^2+16x+96 &= 0&& \text{simplify left side} \\[1 em]-16x^2+17x+101 &= 0&& \\[1 em] \end{aligned} $$
$ -16x^{2}+17x+101 = 0 $ is a quadratic equation.
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