We solve this inequality with a sign chart.
#x^3<=4x^2+3x#
#x^3-4x^2-3x<=0#
#x(x^2-4x-3)<=0#
We need the roots of the quadratic equation
#x^2-4x-3=0#
The discrimenant is
#Delta=b^2-4ac=(-4)^2-4*(1)*(-3)=16+12=28#
As, #Delta>0#, there are 2 real roots
#x_2=(-b+sqrtDelta)/2=1/2(4+sqrt28)=2+sqrt7=4.646#
#x_1=(-b-sqrtDelta)/2=1/2(4+sqrt28)=2-sqrt7=-0.646#
Let #f(x)=x(x-x_1)(x-x_2)#
We can build the sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##x_1##color(white)(aaaa)##0##color(white)(aaaaaa)##x_2##color(white)(aaaa)##+oo#
#color(white)(aaaa)##x_1##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##x_2##color(white)(aaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#
Therefore,
#f(x)<=0# when #x in (-oo, 2-sqrt7] uu [0, 2+sqrt7]#
graph{x^3-4x^2-3x [-12.34, 12.97, -9.01, 3.65]}