Помогите решить пж (1+i)(2+2i)=

X^2 - 5X + 6 = 0 D = 25 - 24 = 1 ; √ D = 1 X1 = ( 5 + 1 ) : 2 = 3 X2 = ( 5 - 1 ) : 2 = 2 Ответ 3 и 2 ---------------------------- X^2 - 7X + 10 = 0 D = 49 - 40 = 9 ; √ D = 3 X1 = ( 7 + 3 ) : 2 = 5 X2 = ( 7 - 3 ) : 2 = 2 Ответ 5 и 2 ----------------------------- Y^2 + 8Y + 16 = 0 D = 64 - 64 = 0 Y = ( - 8 ) : 2 = ( - 4 ) ----------------------------- Y^2 - 10Y + 25 = 0 D = 100 - 100 = 0 Y = 10 : 2 = 5