Respuesta :

davidmaharjan62

Answer:

[tex] {x}^{2} + x = 12 \\ {x}^{2} + x - 12 = 0 \\ {x }^{2} + 4x - 3x - 12 = 0 \\ x(x + 4) - 3(x + 4) = 0 \\ (x + 4)(x - 3) = 0 \\ either \: \: \: or \: \\ x = - 4 \: \: \: \: \: \: \: \: \: \: \: \: x = 3[/tex]

the answer is -4 and 3

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