jmc

First, we write the given equation as $$\sqrt{x}+\sqrt{x+7}+2\sqrt{x^2+7x}+2x=35.$$Let $y=\sqrt{x}+\sqrt{x+7}.$ Then $$y^2=x+2\sqrt{x(x+7)}+x+7=2\sqrt{x^2+7x}+2x+7.$$Hence, $y+y^2-7=35.$ Then $y^2+y-42=0,$ which factors as $(y-6)(y+7)=0.$ Since $y$ is positive, $y=6.$

Hence, $$\sqrt{x}+\sqrt{x+7}=6.$$Then $\sqrt{x+7}=6-\sqrt{x}.$ Squaring both sides, we get $$x+7=36-12\sqrt{x}+x.$$Then $12\sqrt{x}=29,$ so $x={\left(\frac{29}{12}\right)}^2=\boxed{\frac{841}{144}}.$ We check that this solution works.