2

[tex]1)\\\\\begin{cases}\frac{x}{2} > \frac{x-3}{3}\ \ \ \ |\cdot 6\\3(x+1)-5 < 2(x-3)+4\end{cases}\\\\\begin{cases}3x > 2(x-3)\\3x+3-5 < 2x-6+4\end{cases}\\\\\begin{cases}3x > 2x-6\\3x-2x < -6+4-3+5\end{cases}\\\\\begin{cases}3x-2x > -6\\x < 0\end{cases}\\\\\begin{cases}x > -6\\x < 0\end{cases}\\\\x\in(-6;0)[/tex]

[tex]2)\\\\\begin{cases}\frac{3x-1}{4}\le x-2 \ \ \ \ |\cdot 4\\(x-2)x-5\ge x(x+1)-29\end{cases}\\\\\begin{cases}3x-1\le 4x-8\\x^2-2x-5\ge x^2+x-29\end{cases}\\\\\begin{cases}3x-4x\le -8+1\\x^2-2x-x^2-x\ge -29+5\end{cases}\\\\\begin{cases}-x\le -7\ \ \ |:(-1)\\-3x\ge -24\ \ \ |:(-3)\end{cases}\\\\\begin{cases}x\ge7\\ x\le 8\end{cases}\\\\x\in[7;8][/tex]

3.

[tex]\frac{1}{2}(3-x)-3(4-2x) > x-3\ \ \ |\cdot2\\\\3-x-6(4-2x) > 2x-6\\\\3-x-24+12x > 2x-6\\\\-x+12x-2x > -6-3+24\\\\9x > 15\ \ \ |:9\\\\x > \frac{5}{3}\\\\x > 1\frac{2}{3}\\\\x\in(1\frac{2}{3};+\infty)[/tex]

[tex]2 x^{2} + 7x - 12 =0\\\\ a=2 ,\ \ b=7 ,\ \ c=-12\\\\ D = b^2 - 4ac = 7^2 - 4\cdot2\cdot( - 12) = 49 + 96 = 145\\\\\sqrt{D} =\sqrt{145}\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{-7-\sqrt{145}}{2\cdot2}=\frac{-7-\sqrt{145} }{4 }\\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{-7+\sqrt{145}}{2\cdot2}=\frac{-7 +\sqrt{145} }{4}[/tex]