Question 1011021
You know A area of a cardboard (rectangular), u length of side of squares cut from corners, and v volume of the open box created when folding up the flaps.  You want to know the cardboard's dimensions, x and y.


{{{system(A=242,u=2,x=unknownDimension,y=unknownOtherDimension,v=252)}}}


The area of the BASE after folding the flaps up, removes TWO of u at the ends, because two ends for each dimension being folded up to make this height of u.


Height of box, u
Dimensions of the base, x-2u  and y-2u
Box volume, u*(x-2u)(y-2u)
Area of cardboard, A=xy


volume equation for this description:
{{{v=u(x-2u)(y-2u)}}}


Eliminate either variable using A equation; {{{y=A/x}}}.
{{{highlight_green(v=u(x-2u)((A/x)-2u))}}}
Simplify this and solve for x, in pure symbols if possible.


{{{v=u(A-2uA/x-2ux+4u^2)}}}
{{{Au-2u^2A/x-2u^2x+4u^3=v}}}
Multiply members by x to clear the denominator.
{{{Aux-2u^2A-2u^2x^2+4u^3x=vx}}}
{{{-2u^2x^2+Aux+4u^3x-2u^2A=vx}}}
{{{2u^2x^2-Aux-4u^3x+2u^2A=-vx}}}
{{{2u^2x^2+vx-Aux-4u^3x+2u^2A=0}}}
{{{2u^2x^2+(v-Au-4u^3)x+2u^2A=0}}}
or as
{{{highlight_green(2u^2x^2+(v-Au-4u^3)x+2Au^2=0)}}}-----Quadratic equation in the variable x, all in pure symbolic form.


You can try general solution formula for quadratic equation on that but any simplifications to try might be a mess.  SUBSTITUTE THE GIVEN VALUES NOW, AND SIMPLIFY FROM THAT BEFORE FINISHING TO SOLVE FOR x.


The rest is left to you to continue and finish.  Remember, you are looking for dimensions x and y.