Idon't remember what it is that i'm doing, as i did the lessons before spring break. ?

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

1. 0.3 •4 is the same as 4 groups of 0.32. 7•0.8 is the same as 7 groups of 0.83. it increases by 0.5 every time, so 0.5x4 it increases by 0.9 every time, so 0.9x5. 0.9•12=10.86. 0.47•5=2.35i can finish the next ones in a separate post