Question 930177
Let x = smaller number
   x+1= larger number

A number increased by its square is equal to 9 times the next number
x^2 = 9(x+1)
x^2 = 9x+9

x^2-9x-9 = 0

Use Quadratic formula to solve for x
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-(-9) +- sqrt( (-9)^2-4*(1)*(-9) ))/(2*(1)) }}}

{{{x = (9 +- sqrt( 81+36 ))/(2) }}}

{{{x = (9 +- sqrt( 117 ))/(2) }}}

x=9.9083269131959839396788319012057 or 10
or 
x=-0.908326913195983939678831901205 or -1

Substitute x with 10
(10)^2 = 9(10+1)
   100 = 9(11)
   100 = 99
Substitute x with -1
(-1)^2 = 9(-1+1)
    1  = 9(0)
    1  = 0

The number is -1 or 10.

You could also use completing the square to solve for x
x^2-9x-9= 0 
Add 9 on both side
x^2-9x+9= 0+9
x^2-9x  = 9
Divide B coefficient by 2 and square it. Then add it on both side
x^2-9x+81/4=9+81/4
x^2-9x+81/4=36/4+81/4
x^2-9x+81/4=117/4
  (x-9/2)^2=117/4
Square root on both side
√(x-9/2)^2 = √117/4
x-9/2 = 3/2√13
Add 9/2 on both side. Add ± on the right side.
x-9/2+9/2 =9/2±3/2√13
x = 9/2 ±3/2√13
 
x =  9.9083269131959839396788319012057 or 10
or
x = -0.908326913195983939678831901205  or -1