Extraneous solutions, is answers that we get because of squaring both sides of the radical equation, but in reality, they are not going to be the solutions of the given equation.
(√(4x+41))²=(x+5)²
4x+41=x²+10x+25
x²+6x-16=0
(x-2)(x+8)=0
x1=2 , x2=-8,
And now we must to check them by substitution into initial equation
√(4x+41)=x+5
1) x=2, √(4*2+41)=2+5, √49=7, 7=7 true
2) x=-8, √(4*(-8)+41)=-8+5, √9 =-3 false,
so an extraneous solution x=-8