Помогите пожалуйста решить.

Решите задачу:

$a)\ (1-6y)(1+6y)-14\ \textgreater \ 7y(2-5y),\\1^2-(6y)^2-14\ \textgreater \ 14y-35y^2,\\-36y^2+35y^2+1-14-14y\ \textgreater \ 0,\\-y^2-14y-13\ \textgreater \ 0\ |:(-1),\\y^2+14y+13\ \textless \ 0,\\\\y^2+14y+13=0,\\D=14^2-4\cdot1\cdot13=144,\\\\y_{1,2}=\frac{-14\pm\sqrt{144}}{2}=\frac{-14\pm12}{2}=-7\pm6,\\y_1=-7+6=-1,\\y_2=-7-6=-13,\\\\a\ \textgreater \ 0\ \Longrightarrow\ -13\ \textless \ y\ \textless \ -1,\\\\OTBET:\ (-13;\ -1).$

$b)\ (5x-2)+36\ge5x(5x-3),\\5x-2+36\ge25x^2-15x,\\25x^2-15x-5x+2-36\le0,\\25x^2-20x-34\le0,\\\\25x^2-20x-34=0,\\D=(-20)^2-4\cdot25\cdot(-34)=400+3400=3800,\\\\x_{1,2}=\frac{20\pm\sqrt{3800}}{2\cdot25}=\frac{20\pm10\sqrt{38}}{50}=\frac{2\pm\sqrt{38}}{5},\\\\x_1=\frac{2+\sqrt{38}}{5},\\\\x_2=\frac{2-\sqrt{38}}{5},\\\\a\ \textgreater \ 0\ \Longrightarrow\ \frac{2-\sqrt{38}}{5}\le x\le\frac{2+\sqrt{38}}{5}.\\\\OTBET:\ [\frac{2-\sqrt{38}}{5};\ \frac{2+\sqrt{38}}{5}].$