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Solve the following equations:
$4|x-1| - 7 = 13$
$\displaystyle{\left| \frac{x-2}{3} \right| = 4}$
$4x^2 + 8x = -3$
$2x^2 = 5x + 12$
$\sqrt{33-2x} = x+1$
$7 = \sqrt{39 + 3x} - x$
$3^x + 2 \cdot 3^{x} = 1$
$\sqrt{x-5} + \sqrt{x} = 5$
$\sqrt{x + \sqrt{x-3}} = 3$
$x^3 + 6x^2 = 2x + 12$
$2x - 5\sqrt{x} + 2 = 0$
$3x^{2/3} - 5x^{1/3} - 2 = 0$
$2x^{-2} - 11x^{-1} + 5 = 0$
$\log_2 x^2 = \log_2^2 x$
$\log_3 (x + 6) = 3 - \log_3 x$
$\log_4 (x+12) - \log_4 (x-3) = 2$
$\frac{1}{2} \log_4 (2x+1) = \log_4 3$
$\log_2 (x+5) - \log_2 x = \log_2 4$
$\log_4 (x+3) + \log_4 (x-3) = 2$
$\displaystyle{\frac{2}{x-5} - \frac{1}{x+3} = \frac{8}{(x+3)(x-5)}}$