Atriangle has a vertical height 2cm longer than it's base lengths. it's area is 20cm^2. find the base and the height of the triangle.

Let the base be b.  The the height of the triangle is b+2.
                                                   (base)(height)
The area of any triangle is A =
                                                           2
                                    (b)(b+2)
Here, A = 20 cm^2 =
                                          2

so 40 cm^2 = b^2 + 2b, or     b^2 + 2b - 40 = 0

solving this quadratic with the quadratic formula:

         -2 plus or minus sqrt( 4-4(1)(-40) )
b =
                                   2

   = -1 plus or minus sqrt( 4 + 160)  = -1 plus or minus sqrt 164

Note that sqrt(164) = sqrt(4)sqrt(41) = 2sqrt(41)

so b = -1 plus or minus 2sqrt(41).  We must eliminate the negative root here.  So, we get   b = -1 + sqrt(41).  Then the height is h = -1 + 2 + sqrt(41), or

b = -1 + sqrt(41)    and     h = b+2  = 1 + sqrt(41)          (answer)