Дана геометрическая прогрессия. (0 Первый член 2x Второй x-5 Третий (x-1)/3 Надо найти q....

$b_{1}=2x\\\\b_{2}=x-5}\\\\b_3} =\frac{x-1}{3}\\\\\\q=\frac{b_{2}}{b_{1}}=\frac{x-5}{2x}\\\\q=\frac{b_{3}}{b_{2}} = \frac{x-1}{3(x-5)}\\\\\\\frac{x-5}{2x} =\frac{x-1}{3(x-5)}\\\\\frac{x-5}{2x}-\frac{x-1}{3x-15} =0\\\\\frac{3x^{2}-15x-15x+75-2x^{2}+2x}{6x(x-5)}=0\\\\\frac{x^{2}-28x+75}{x(x-5)}=0\\\\x^{2}-28x+75=0; x\neq 0 ;x\neq5\\\\x_{1}=25;x_{2} =3\\\\ q _{1}=\frac{25-5}{2*25}=\frac{20}{50}=\frac{2}{5}$ \\\\ $q_{2}= \frac{3-5}{2*3}=- \frac{1}{3}$