40-14x+x^2=2(x-4)√x.

Решите задачу:

$ODZ:x\geq0\\\\40-14x+x^{2} =2(x-4)\sqrt{x} \\\\x^{2} -14x+40=2\sqrt{x}(x-4)\\\\(x-4)(x-10)-2\sqrt{x}(x-4)=0\\\\(x-4)(x-10-2\sqrt{x})=0\\\\\left[\begin{array}{ccc}x-4=0\\x-10-2\sqrt{x} =0\end{array}\right\\\\1)x-4=0\Rightarrow x_{1}=4\\\\2)x-10-2\sqrt{x}=0\\\\\sqrt{x}=m,m\geq0\\\\m^{2} -2m-10=0\\\\D=(-2)^{2}-4*(-10)=4+40=44=(2\sqrt{11})^{2}\\\\m_{1}=\frac{2+2\sqrt{11}}{2}=1+\sqrt{11}\\\\m_{2}=\frac{2-2\sqrt{11}}{2}=1-\sqrt{11}$

$\sqrt{x}=1+\sqrt{11}\\\\x_{2} =(1+\sqrt{11})^{2}\\\\x_{2} =1+2\sqrt{11}+11=12+2\sqrt{11}\\\\Otvet:\boxed{4;12+2\sqrt{11}}$