HMMT February

$(2 x+1)(3 x+1)(5 x+1)(30 x+1)=[(2 x+1)(30 x+1)][(3 x+1)(5 x+1)]=\left(60 x^{2}+32 x+1\right)\left(15 x^{2}+8 x+1\right)=(4 y+1)(y+1)=10$, where $y=15 x^{2}+8 x$. The quadratic equation in $y$ yields $y=1$ and $y=-\frac{9}{4}$. For $y=1$, we have $15 x^{2}+8 x-1=0$, so $x=\frac{-4 \pm \sqrt{31}}{15}$. For $y=-\frac{9}{4}$, we have $15 x^{2}+8 x+\frac{9}{4}$, which yields only complex solutions for $x$. Thus the real solutions are $\frac{-4 \pm \sqrt{31}}{15}$.