Đặt x^2 +x = y
ta có pt : y^2 + 4y - 12 = 0
=> y^2 + 6y - 2y - 12 = 0
=> y(y + 6) - 2( y + 6) = 0
=> ( y - 2)( y + 6) = 0
=> y = 2 ; y =- 6
(+) y = 2 => x^2 + x = 2
=> x^2 +x - 2 = 0 => x^2 + 2x - x - 2 = 0
=> x( x+ 2) - (x+2) = 0
=> ( x - 1)( x+ 2) = 0
=> x = 1 ; x = - 2
(+) x^2 + x = -6 => x^2 + x + 6 = 0
Vì x^2 + x+ 6 = x^2 + 2x.1/2 + 1/4 + 23/4 = ( x + 1/2)^2 +23/4 > 0
=> pt vô nghiệm