Solve and discussed by the values of real m this equation:
$\displaystyle m^2 x^4 + 18mx^3 + \left( {2m - 175} \right)x^2 - 78x - 8 = 0$
The equation can be written as
$\displaystyle m^2x^4+(18x^3+2x^2)m-175x^2-78x-8=0$
$\displaystyle \Delta=b^2-4ac=4x^4(16x+3)^2$
$\displaystyle m_1=-\frac{25}{x}-\frac{4}{x^2}, \ m_2=\frac{7}{x}+\frac{2}{x^2}$
Then the equation can be factorize:
$\displaystyle x^4\left(m+\frac{25}{x}+\frac{4}{x^2}\right)\left( m-\frac{7}{x}-\frac{2}{x^2}\right)=0$
or $\displaystyle (mx^2+25x+4)(mx^2-7x-2)=0$
Now you have to solve and discuss two quadratic equations:
$\displaystyle mx^2+25x+4=0$ and $\displaystyle mx^2-7x-2=0$.
Can you do that?