[tex] {x}^{2} + 8x + 12 = 0 \\ = > {x}^{2} + 6x + 2x + 12 = 0 \\ = > x(x + 6) + 2(x + 6) = 0 \\ = > (x + 2)(x + 6) = 0 \\ = > x = - 2 \: \: or \: \: - 6[/tex]
[tex]3x + 6 = 0 \\ = > 3x = - 6 \\ = > x = \frac{ - 6}{3} \\ = > x = - 2[/tex]
Here, the common solution is -2.
Answer:
x = -2.
Hope you could get an idea from here.
Doubt clarification - use comment section.
