Question 709295
The original area is A = xy


The new area is y(x+3) = xy + 3y and this is equal to 20, so


xy + 3y = 20


We know that the new figure is a square, so y = x+3


Plug this in and solve for x 


xy + 3y = 20


x(x+3) + 3(x+3) = 20


x^2 + 3x + 3x + 9 = 20


x^2 + 6x + 9 - 20 = 0


x^2 + 6x - 11 = 0


Now use the quadratic formula to solve for x



{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(6)+-sqrt((6)^2-4(1)(-11)))/(2(1))}}} Plug in {{{a = 1}}}, {{{b = 6}}}, {{{c = -11}}}


{{{x = (-6+-sqrt(36-(-44)))/(2)}}}


{{{x = (-6+-sqrt(36+44))/(2)}}}


{{{x = (-6+-sqrt(80))/2}}}


{{{x = (-6+sqrt(80))/2}}} or {{{x = (-6-sqrt(80))/2}}}


{{{x = (-6+4*sqrt(5))/2}}} or {{{x = (-6-4*sqrt(5))/2}}}


{{{x = -3+2*sqrt(5)}}} or {{{x = -3-2*sqrt(5)}}}


{{{x = 1.472136}}} or {{{x = -7.472136}}}


Note: Solutions above are approximate


Ignore the negative solution


So x is roughly 1.472136 and y = x+3 = 1.472136+3 = 4.472136


So x = 1.472136 and y = 4.472136



The area of the original rectangle is therefore


A = xy


A = 1.472136*4.472136


A = 6.583592402496


So the area of the original rectangle is roughly 6.583592402496 square units.