Answer:
The answer is B.
Step-by-step explanation:
You have make y equals to 0 and solve each and every equation one by one :
Option A,
[tex] {x}^{2} + x - 6 = 0 \\ {x}^{2} + 3x - 2x - 6 = 0 \\ x(x + 3) - 2(x + 3) = 0 \\ (x - 2)(x + 3) = 0 \\ x = 2 \: or \: - 3[/tex]
Option B,
[tex] {x}^{2} - x - 6 = 0 \\ {x}^{2} - 3x + 2x - 6 = 0 \\ x(x - 3) + 2(x - 3) = 0 \\ (x + 2)(x - 3) = 0 \\ x = - 2 \: or \: 3[/tex]
Option C,
[tex] {x}^{2} - 5x - 6 = 0 \\ {x}^{2} + x - 6x - 6 = 0 \\ x(x + 1) - 6(x + 1) = 0 \\ (x - 6)(x + 1) = 0 \\ x = 6 \: or \: - 1[/tex]
Option D,
[tex] {x}^{2} + 5x - 6 = 0 \\ {x}^{2} - x + 6x - 6 = 0 \\ x(x - 1) + 6(x - 1) = 0 \\ (x + 6)(x - 1) = 0 \\ x = - 6 \: or \: 1[/tex]
