(x+4)*(x-3)+(x-5)*(x+4)=0, х²-3х+4х-12+х²+4х-5х-20=0, 2х²-32=0, D = b 2 - 4ac = 256, D > 0 => 2 вещественных решения, √D = 16 x 1 = -b + √D/2а = 0 + 16/2 × (2) = 4 x 2 = -b - √D/2а = 0 - 16/2 × (2) = -4