Пожалуйсто, ооочень срочно!

Решите задачу:

$C_{x}^{3}=\frac{x!}{3!*(x-3)!}=\frac{(x-3)!*(x-2)(x-1)*x}{1*2*3*(x-3)!}=\frac{x(x-1)(x-2)}{6}\\\\\frac{x(x-1)(x-2)}{6}=\frac{5x(x-3)}{4}\\\\4(x-1)(x-2)=30(x-3)\\\\4x^{2} -12x+8-30x+90=0\\\\4x^{2}-42x+98=0\\\\2x^{2}-21x+49=0\\\\D=(-21)^{2}-4*2*49=441-392=49=7^{2}\\\\x_{1} =\frac{21-7}{4}=3,5-neyd\\\\x_{2}=\frac{21+7}{4}=7\\\\Otvet:\boxed{7}$