1) (x+3)^2=2x+6 2) 2x^2-8=0

Question

1) (x+3)^2=2x+6

2) 2x^2-8=0

Answer 1

1) (x+3)^2=2x+6
х^2 + 6х + 9 - 2х - 6 = 0
х^2 + 4х + 3 = 0
По теореме Виета:  х1 = -1
                                  х2 = -3

2) 2x^2-8=0
2x^2 = 8
x^2 = 4
х = 2