Question 396120
Assuming the 4 and 2 are exponents,
{{{m^4 + 13m^2 - 48 = 0}}}
Sub x for m^2
{{{x^2 + 13x - 48 = 0}}}
(x+16)(x-3) = 0
x = 3, -16
----------
Sub back
{{{m^2 = 3}}}
m = ± sqrt(3)
------------
{{{m^2 = -16}}}
m = ± 4i
================
{{{m^4 + 13m^2 - 48 = 0}}}
You might factor that, too.
{{{m^2 + 16)(m^2 - 3) = 0}}}
Same answers, simpler, but if factoring isn't possible, substitution always works.