Сократить дробь. 6x²-7x-3/2x²-x-3=...?

$\frac{6x^2-7x-3}{2x^2-x-3},\\\\1)\ 6x^2-7x-3=6(x-x_1)(x-x_2)\\6x^2-7x-3=0,\\D=7^2+6\bullet3\bullet4=121,\\\\x_1=\frac{7+\sqrt{121}}{6\bullet2}=\frac{3}{2},\ \ x_2=\frac{7-11}{12}=-\frac{1}{3},\\\\6x^2-7x-3=6(x-\frac{3}{2})(x+\frac{1}{3});\\\\\\2)\ 2x^2-x-3=2(x-x_1)(x-x_2),\\2x^2-x-3=0,\\D=25,\\\\x_1=\frac{1+5}{4}=\frac{3}{2},\ \ x_2=\frac{1-5}{4}=-1,$
$2x^2-x-3=2(x-\frac{3}{2})(x+1);\\\\\\\frac{6x^2-7x-3}{2x^2-x-3}=\frac{6(x-\frac{3}{2})(x+\frac{1}{3})}{2(x-\frac{3}{2})(x+1)}=\frac{3(x+\frac{1}{3})}{x+1}=\frac{3x+1}{x+1}.$

Ответ: $\frac{3x+1}{x+1}.$