Помогите решить 1123(а,в)

Решите задачу:

$\frac{12x^2+1}{2x+3}=4x-1 \\ 12x^2+1=(4x-1)(2x+3) \\ 12x^2+1=8x^2+12x-2x-3 \\ 12x^2+1=8x^2+10x-3 \\ 12x^2+1-8x^2-10x+3=0 \\ 4x^2-10x+4=0 |:4 \\ x^2-2,5x+1=0 \\ D=b^2-4ac \\ D=2,5^2-4*1*1=6,25-4=2,25 \\ x_{1,2} = \frac{2,5+/- \sqrt{2,25} }{2} \\ x_1 = \frac{2,5+1,5}{2} = \frac{4}{2} = 2 \\ x_2 = \frac{2,5-1,5}{2} = \frac{1}{2} = 0,5 \\ \\ 3y+ \frac{2y^2+y}{4y+10}=-3 \\ \frac{3y(4y+10)+(2y^2+y)}{4y+10} = -3 \\ \frac{12y^2+30y+2y^2+y}{4y+10}=-3 \\ \frac{14y^2+31y)}{4y+10}=-3 \\$
$14y^2+31y=-3(4y+10) \\ 14y^2+31y=-12y-30 \\ 14y^2+31y+12y+30=0 \\ 14y^2+43y+30=0 \\ D=43^2-4*30*14=169 \\ x_{1,2} \frac{-b+/- \sqrt{D} }{2a} \\ x_{1,2} \frac{-43+/-13}{28} \\ x_1 = \frac{-43+13}{28} = \frac{-30}{28} \\ x_2 = \frac{-43-13}{28} = \frac{-56}{28} = -2 \\ \\ x_1 = -1\frac{1}{14} ; x_2 = -2$