2017/022) Solve $(2x+1)(30x+1)(3x+1)(5x+1) = 10$

we have $(2x+1)(30x+1)(3x+1)(5x+1) = 10$
or
$(60x^2+ 32x + 1)(15x^2+ 8x + 1) = 10$

letting $15x^2 + 8x = t$
$(4t+1)(t+1) = 10$
or
$4t^2 + 5 t + 1 = 10$
or $4t^2 + 5t - 9 = 0$
or $(4t+9)(t-1) = 0$

t =  1 or -9/4
t = 1 gives
$15x^2 + 8x-1=0$ giving $x = \dfrac{-4\pm\sqrt{31}}{15}$

or  $(15x^2 + 8x +\frac{9}4{4}) = 0$
or $(60x^2+ 32x + 9) = 0$
this gives complex solution
so solutions are  $x = \dfrac{-4\pm\sqrt{31}}{15}$