Answer
The solutions are -4 and 3.
Work Step by Step
(y-2)(y+3)=6
$y^{2}$+3y-2y-6=0
$y^{2}$+y-6+6=6+6
$y^{2}$+y=12
$y^{2}$+y-12=0
$y^{2}$+4y-3y-12=0
y(y+4)-3(y+4)=0
(y-3)(y+4)=0
y-3=0 or y+4=0
y=3 or y=-4
The solutions are -4 and 3.
Check
Let x=-4
(y-2)(y+3)=6
(-4-2)(-4+3)=6
-6(-1)=6
6=6
Let x=3
(y-2)(y+3)=6
(3-2)(3+3)=6
1*6=6
6=6
