Question

Solve the equation: $(y^2+5y)(y^2+5y-2)-24=0$

Answer 1

$(y^2+5y)(y^2+5y-2)-24=0$
Let, $y^2+5y=a$
$a(a-2)-24=0$
$a^2-2a-24=0$
$a^2-6a+4a-24=0$
$a(a-6)+4(a-6)=0$
$(a-6)(a+4)=0$
$a-6=0$ or $a+4=0$
By resubstituting,
$y^2+5y=a$
$y^2+5y-6=0$ or $y^2+5y+4=0$
$y^2+6y-y-6=0$ or $y^2+4y+y+4=0$
$y(y+6)-1(y+6)=0$ or $y(y+4)+1(y+4)=0$
$(y+6)(y-1)=0$ or $(y+4)(y+1)=0$
$y-1=0$ or $y+6=0$ or $y+4=0$ or $y+1=0$
$\therefore y=1$ or $y=-6$ or $y=-4$ or $y=-1$
$y=-6,-4,-1,1$.