Two positive numbers differ by 3 and the sum of their squares is 120, Find their numbers:

Having a hard time figuring this question out, pretty sure its a simultaneous equation but cant get the two numbers :(, if anyone can help please.

x^2 + (x-3)^2= 120
x^2 + x^2-6x+9=120
2x^2-6x+9=120
2x^2-6x=111
x^2-3x=55.5
(x-1.5)^2=55.5 + (9/4)
X= sqrt (55.5+ 2.25) +1.5
X= sqrt 57.75 + 1.5
X= 9.099

X-3= 6.099

The two numbers are 6.099 and 9.099.

Let x = one of the numbers so then x+3 is the other number.

x^2 + (x+3)^2 = 120
x^2 + x^2 + 6x + 9 = 120
2x^2 + 6x - 111 = 0

solve for x, use the quadratic formula