Question

Решить уравнение; 1) x^2+6x-7=0; 2) x^2+12x+32=0

Answer 1

x²-6x-7=0

D=36-(-7)*4=36+28=64

x1=(6+√64)/2= (6+8)/2=7

x2=(6-√64)/2=(6-8)/2=-1

x²+12x+32=0

D=12²-4*32=144-128=16

x1=(-12+√16)/2=(-12+4)/2=-4

x2=(-12-√16)/2=(-12-4)/2=-8