Помогите решить четыре уравнения..Очень прошуу..

Решите задачу:

$\frac{(2x+1)^2}{25} + \frac{x-1}{3} =x \\\ 3(2x+1)^2+25(x-1)=75x \\\ 3(4x^2+4x+1)+25x-25-75x=0 \\\ 12x^2+12x+3+25x-25-75x=0 \\\ 12x^2-38x-22=0 \\\ 6x^2-19x-11=0 \\\ D=361+264=625 \\\ x_1= \frac{19+25}{12} = \frac{44}{12}= \frac{11}{3} \\\ x_2= \frac{19-25}{12} = -\frac{6}{12}= -0.5$

$\frac{(3x+2)^2}{11} - \frac{x+5}{4} =x^2 \\\ 4(3x+2)^2-11(x+5)=44x^2 \\\ 4(9x^2+12x+4)-11x-55-44x^2=0 \\\ 36x^2+48x+16-11x-55-44x^2=0 \\\ -8x^2+37x-39=0 \\\ 8x^2-37x+39=0 \\\ D=1369-1248=121 \\\ x_1= \frac{37+11}{16}=3 \\\ x_2= \frac{37-11}{16}=\frac{26}{16}=\frac{13}{8}$

$\frac{(2-x)^2}{3} - 2x=\frac{(7+2x)^2}{5} \\\ 5(2-x)^2-30x=3(7+2x)^2 \\\ 5(4-4x+x^2)-30x=3(49+28x+4x^2) \\\ 20-20x+5x^2-30x-147-84x-12x^2=0 \\\ -7x^2-134x-127=0 \\\ 7x^2+134x+127=0 \\\ D_1=4489-889=3600 \\\ x_1= \frac{-67-60}{7} =-\frac{-127}{7} \ \\\ x_2= \frac{-67+60}{7} =-1$

$\frac{(6-x)^2}{8} +x=7- \frac{(2x-1)^2}{3} \\\ 3(6-x)^2+24x=168-8(2x-1)^2 \\\ 3(36-12x+x^2)+24x=168-8(4x^2-4x+1) \\\ 108-36x+3x^2+24x-168+32x^2-32x+8 \\\ 35x^2-44x-52=0 \\\ D_1=484+2030=2304 \\\ x_1= \frac{22+48}{35} =2 \\\ x_2= \frac{22-48}{35} =- \frac{26}{35}$