[@dan-florinchereches4892]

x=y+4 ;y not 0 or -1
y+4=√(27/y)-(20/(y+1))
y+4=√(27y+27-20y)/(y*(y+1))
y+4=√(7y+27)/(y(y+1)) squaringand multiplying
(y+4)^2*y*(y+1)=7y+27
(y^2+8y+16)(y^2+y)=7y+27
Y^4+9y^3+24y^2+16y=7y+27
y^4+9y^3+24y^2+9y-27=0
(y^2+Ay+3)(y^2+By-9)=0
A+B=9 AB-6=24 B-3A=3 false
Or
(y^2+Ay-3)(y^2+By+9)=0
A+B=9 AB+6=24 3A-B=3 A=3;B=6
Or
(y^2+Ay-1)(y^2+By+27)=0
So (y^2+3y+3)(y^2+6y-9)=0
=>((y+1.5)^2-2.25+3)((y+3)^2-18)=0
So y=3+-3√2
=> x=7+3√2 or x=7-3√2

[@superacademy247]

Awesome 💯👍

[@dan-florinchereches4892]

I messed up the algebra should have gotten when substituting back
(y^2+3y-3)(y^2+6y+9)=0
Then ((y+3/2)^2-21/4)(y+3)^2=0 leads to solutions in the video

[@遠傳五華]

"70% Failed to Attain PassMark Score" might not be true, as no trap in the quiz.