In rectangle ZACH, ZC=x^2-12 and AH=6x-5, find x.?

Assuming that 'Z' is the top left hand corner and 'C' is the bottom left, corner,and 'A' and 'H' are the same [but on the right hand side] the both equations will be equal.

so

6x - 5 = x² - 12

x² - 12 - 6x + 5 = 0 ---> x² - 6x - 7 = 0

x² - 7x + x - 7 = 0

x(x - 7) + 1(x -7) = 0

x = -1 and x = 7

Value of 'x' can't be negative so answer is 7

Since it is a rectangle, the two diagonals must be equal. So

x^2 - 12 = 6x - 5

x^2 - 6x - 7 = 0

So we have a simple quadratic equation.

I can see that it factorises so

(x - 7)(x + 1) = 0

So

x - 7 = 0

x = 7

or

x + 1 = 0

x = -1

This answer would give negative values of the lenghts, so

x = 7 is the correct answer.

Check

7^2 - 12 = 49 - 12 = 37

(6 * 7) - 5 = 42 - 5 = 37

diagonals in rectangle are equal

x^2 -12 =6x-5

x^2 -6x -7 =0

(x-7)(x+1) = 0

x=7 or x = - 1.......> rejected

so x = 7...................................answer