Answer
The lengths of the sides are $21$, $19$ and $8$
Work Step by Step
Since the perimeter is equal to the sum of all sides, we have that:
$x^2+3+4x+5+2x=48$
$x^2+6x+8=48$
$x^2+6x+8-48=0$
$x^2+6x-40=0$
$x^2+10x-4x-40=0$
$x(x+10)-4(x+10)=0$
$(x-4)(x+10)=0$
Therefore the solutions are $x_{1}=4$ and $x_{2}=-10$, but since the sides can not have a negative the value, we discard $x_{2}=-10$
So the sides lengths of the sides are:
$x^2+3=(4)^2+3=16+3=19$
$4x+5=4(4)+5=16+5=21$
$2x=2(4)=8$
