the complete question is
The sum of the square of 2 number is 34. The first number is one less than twice the 2nd number. Find the number
Let
2x-1-------> the first number
x------->the second number
we know that
[tex] (2x-1)^{2} +x^{2} =34\\ 4x^{2} -4x+1+x^{2} =34\\ 5x^{2} -4x-33=0\\ 5x^{2} -15x+11x-33=0\\ 5x*(x-3)+11*(x-3)=0\\ (5x+11)*(x-3)=0 [/tex]
therefore
[tex] x=3\\ 2x-1=5 [/tex]
the answer is
the numbers are
[tex] 3,5 [/tex]
