Question

Срочно плииииииз!!!!

Answer 1

по теореме Виета:

$x^2+bx+c=0\\\left \{ {{x_1+x_2=-b} \atop {x_1x_2=c}} \right.$

$x^2+12x+6=0\\\left \{ {{x_1+x_2=-12} \atop {x_1x_2=6}} \right.$

$(x_1+x_2)^2=(-12)^2\\x_1^2+2x_1x_2+x_2^2=144\\\\x_1^2+x_2^2=144-2x_1x_2\\x_1^2+x_2^2=144-2*6\\x_1^2+x_2^2=144-12\\x_1^2+x_2^2=132$