Question 983189
A mistake is in this work.  I have not found it yet.


"length of a rectangular piece sheet metal is 3 cm less than twice. "


Twice what?  Width?


L, length
w, WIDTH
c, side of the square cut from each corner
v, volume of result box


These are the facts in the description:
{{{system(L=-3+2w,c=4,v=208)}}}


You also know that {{{wLc=v}}}.


MAKE SOME EQUATIONS OR EXPRESSIONS

A drawing will help you  (but not shown here).


Base area,  {{{(L-2c)(w-2c)}}}
Volume, {{{c(L-2c)(w-2c)=v}}}





Simplify some first (doing purely in symbols as far as comfortable).
{{{c(wL-2cw-2cL+4c^2)=v}}}
{{{c(wL-2cL-2cw+4c^2)=v}}}
{{{c(w(2w-3)-2c(2w-3)-2cw+4c^2)=v}}}, substituted for L
{{{c(2w^2-3w-4cw+6c-2cw+4c^2)=v}}},  
{{{c(2w^2-3w-4w-2cw+6c+4c^2)=v}}}
{{{c(2w^2-(3+4+2c)w+6c+4c^2)=v}}}
{{{c(2w^2-(7+2c)w+6c+4c^2)=v}}}
{{{highlight_green(2w^2-(7+2c)w+6c+4c^2-v/c=0)}}}-------You can continue this purely in symbols, or you can substitute NOW for c and for v, and continue the steps from that.  You want to then solve for w, either through factoring or through formula for general solution of quadratic formula.


Continueing from the substitution for c and v and simplifying, you should obtain  {{{highlight_green(2w^2-15w+36=0)}}}.
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You continue this.


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BE CAREFUL.  UNKNOWN MISTAKE IN THE ABOVE WORK.


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<b>Reworking separately on paper,</b>  still not know where mistake is in the above work,  These are two steps found  (some steps not included here):
{{{2w^2-(3+6c)w+6c+4c^2-v/c=0}}}


Substituting for c and v,
{{{2w^2-27w+36=0}}}
Factorable!


{{{highlight((2w-3)(w-12)=0)}}}
From which you can reject the first factor, knowing that the value or w from it is impossible.  Accept the second factor for {{{w=12}}}.  Now, go back and find the corresponding L.