$$6x\cdot3+2x\cdot2-x+14-(x\cdot38x\cdot2-x-3) = 0$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 11 }{ 76 }-\dfrac{ 3 \sqrt{ 157}}{ 76 } & x_2 = \dfrac{ 11 }{ 76 }+\dfrac{ 3 \sqrt{ 157}}{ 76 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 6x\cdot3+2x\cdot2-x+14-(x\cdot38x\cdot2-x-3) &= 0&& \text{simplify left side} \\[1 em]18x+4x-x+14-(76x^2-x-3) &= 0&& \\[1 em]21x+14-(76x^2-x-3) &= 0&& \\[1 em]21x+14-76x^2+x+3 &= 0&& \\[1 em]-76x^2+22x+17 &= 0&& \\[1 em] \end{aligned} $$
$ -76x^{2}+22x+17 = 0 $ is a quadratic equation.
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