Answer:
pythagoras theorem
(2x+1)²= 12²+(x-1)²
4x²+4x+1 = 144+x²-2x+1
4x²+4x+1=x²-2x+145
4x²+4x+1-x²+2x-145=0
3x²+6x-144=0
b²-4ac= 6²-4(3*-144)=36+1728=1764
Δ=1764
Δ is strictly positive, the equation 3x²+6x−144=0 admits two solutions.
(-b-√Δ)/2a = (-6-42)/6=-8
(-b+√Δ)/2a = (-6+42)/6=6
a length is not negative
→x=6