Answer:
The answer is -6 or 2
Step-by-step explanation:
[tex] {x}^{2} + 4x - 12 = 0 \\ 4 \div 2 = 2 \\ x1 = - 2 - u \\ x2 = - 2 + u \\ x1x2 = ( - 2 - u)( - 2 + u) = - 12 \\ 4 - {u}^{2} = - 12 \\ 4 + 12 = {u}^{2} \\ 16 = {u}^{2} \\ \sqrt{16} = + \: or - \sqrt{ {u}^{2} } \\ u = + or - 4 \\ [/tex]
[tex]x1 = - 2 - 4 = - 6 \\ x2 = - 2 + 4 = 2[/tex]
