Question

(Х^2+2х)^2-2(х^2+2х)-3=0

Answer 1

$( {x}^{2} + 2x) {}^{2} - 2( {x}^{2} + 2x) - 3 = 0 \\ t {}^{2} - 2t - 3 = 0 \\ d = {b}^{2} - 4ac \\ d = ( - 2) {}^{2} - 4 \times 1 \times ( - 3) = 4 + 12 = 16 \\ x12 = \frac{ - b + - \sqrt{d} }{2a} \\ x12 = \frac{2 + - 4}{2} \\ x1 = \frac{2 + 4}{2} = \frac{6}{2} = 3 \\ x2 = \frac{2 - 4}{2} = - \frac{2}{2} = - 1 \\ x3 = 1$