## Algebra: A Combined Approach (4th Edition)

The lengths of the sides are $21$, $19$ and $8$
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: $x^2+3=(4)^2+3=16+3=19$ $4x+5=4(4)+5=16+5=21$ $2x=2(4)=8$