One number is six more than the other, so the numbers are x and x+6.
The sum of their squares is 90.
\(x^2+(x+6)^2=90 \\
x^2+x^2+12x+36=90 \\
2x^2+12x+36=90 \ \ \ \ \ \ \ \ \ \ \ |-90 \\
2x^2+12x-54=0 \ \ \ \ \ \ \ \ \ \ \ \ \ |\div 2 \\
x^2+6x-27=0 \\
x^2+9x-3x-27=0 \\
x(x+9)-3(x+9)=0 \\
(x-3)(x+9)=0 \\
x-3=0 \ \lor \ x+9=0 \\
x=3 \ \lor \ x=-9 \\ \\
\hbox{for x=3: } \\
x+6=3+6=9 \\ \\
\hbox{for x=-9: } \\
x+6=-9+6=-3\)
The numbers are: 3 and 9 or -9 and -3.
