Здравствуйте помогите пожалуйста решить задачу фото нужно развернуть даю 100 баллов ​

[tex]\displaystyle\bf\\\frac{150}{x} +\frac{160}{x+30} =5\\\\\\\frac{150}{x} +\frac{160}{x+30} -5=0\\\\\\\frac{150\cdot(x+30)+160\cdot x-5\cdot x\cdot(x+30)}{x\cdot(x+30)} =0\\\\\\\frac{150x+4500+160x-5x^{2} -150x}{x\cdot(x+30)} =0\\\\\\\frac{5x^{2} -160x-4500}{x\cdot(x+30)}=0\\\\\\\left\{\begin{array}{ccc}5x^{2} -160x-4500=0\\x\neq 0\\x\neq -30\end{array}\right\\\\\\5x^{2} -160x-4500=0 \ |: 5\\\\x^{2} -32x-900=0\\\\D=(-32)^{2} -4\cdot (-900)=1024+3600=4624=68^{2}[/tex]

[tex]\displaystyle\bf\\x_{1} =\frac{32+68}{2} =50\\\\\\x_{2} =\frac{32-68}{2} =-18 < 0 \ -neyd\\\\\\Otvet \ : \ V=50[/tex]