Question 686696
The length of a rectangle is 1cm more than twice it's width and 2cm less than the length of the diagonal of the rectangle. Find the dimensions of the rectangle. 
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let x=width
length=2x+1
diagonal=√(width^2+length^2) (by the pythagorean theorem)
=√[(x^2+(2x+1)^2]
=√[x^2+4x^2+4x+1]
=√(5x^2+4x+1)
..
(2x+1)+2=diagonal
2x+3=√(5x^2+4x+1)
square both sides
4x^2+12x+9=5x^2+4x+1
x^2-8x-8=0
solve for x by following quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=1, b=-8, c=-8
ans: 
x≈8.9
2x+1≈18.8
width≈8.9 cm
length≈18.8 cm