Question 1197235
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L = length
W = width


P = perimeter of a rectangle
P = 2(L+W) = 2L+2W


Replace P with the stated perimeter 78 and let's isolate L
P = 2L+2W
78 = 2L+2W
78/2 = 2L/2+2W/2
39 = L+W
L = 39-W


A = area of a rectangle
A = LW
A = (39-W)W
A = -W^2+39W


Plug in the stated area A = 395
I'll replace each W with x
A = -W^2+39W
395 = -x^2+39x
395+x^2-39x = 0
x^2-39x+395 = 0


Use the quadratic formula to solve for x. 
Use a = 1, b = -39, c = 395.
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(-39)+-sqrt((-39)^2-4(1)(395)))/(2(1))}}}


{{{x = (39+-sqrt(1521-1580))/(2)}}}


{{{x = (39+-sqrt(-59))/(2)}}}
The discriminant {{{d = b^2 - 4ac = (-39)^2-4(1)(395) = -59}}} is negative, so there are no real number solutions.
The two roots of x^2-39x+395 = 0 are complex numbers in the form a+bi where {{{i = sqrt(-1)}}}
If your teacher hasn't covered complex or imaginary numbers just yet, then ignore this subsection. 
The summary is that x^2-39x+395 = 0  has no real solutions.


Since x^2-39x+395 = 0 has no real solutions, this means we <u>cannot</u> have a rectangle with perimeter 78 meters and area 395 square meters.
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