Given the differential equation
[tex]\dfrac{dy}{dx} = 12x^2 - 8x[/tex]
integrating both sides with respect to [tex]x[/tex] yields
[tex]\displaystyle \int \frac{dy}{dx} \, dx = \int (12x^2 - 8x) \, dx[/tex]
[tex]y = 4x^3 - 4x^2 + C[/tex]
Use the given point to solve for the constant [tex]C[/tex] :
[tex]-46 = 4(-2)^3 - 4(-2)^2 + C \implies C = 2[/tex]
Then the equation of the curve is
[tex]\boxed{y = 4x^3 - 4x^2 + 2}[/tex]
On the off-chance you instead meant something like
[tex]\dfrac{dy}{dx} = \dfrac{12x^2}{2 - 8x} = -\dfrac{3x}2 - \dfrac38 + \dfrac3{8(1-4x)}[/tex]
integrating would instead give
[tex]y = -\dfrac{3x^2}4 - \dfrac{3x}8 - \dfrac3{32} \ln|1-4x| + C[/tex]
Then
[tex]-46 = -\dfrac94 - \dfrac3{32} \ln(9) + C \implies C = \dfrac3{32}\ln(9) - \dfrac{175}4[/tex]
and the particular solution follows. (But I suspect this is *not* what you meant.)
