Answer:
Option D)Neither solution is extraneous.
Step-by-step explanation:
we have

we know that
two possible solutions are x=-7 and x=1
<u><em>Verify each solution</em></u>
Substitute each value of x in the expression above and interpret the results
1) For x=-7


----> is true
therefore
x=-7 is not a an extraneous solution
2) For x=1


----> is true
therefore
x=1 is not a an extraneous solution
therefore
Neither solution is extraneous
SA= bh+2ls+lw
SA= 26*10 +40*26(2)+40*10
SA=260+1040(2)+400
SA=260+2080+400
SA=2740