(x²+5x)²-4(x²+5x)-12=0

Пусть t=(x²+5x) Тогда t²-4t-12=0 D=16-4*1*(-12)=64 t₁=(4+8)/2=6 t₂=(4-8)/2=-2 Делаем обратную замену: x²+5x=6 x²+5x-6=0 D=25-4*1*(-6)=49 x₁=(-5-7)/2=-6 x₂=(-5+7)/2=1 x²+5x=-2 x²+5x+2=0 D=25-4*1*2=17 x₁=(-5-√17)/2 x₂=(-5+√17)/2

t = x^2 + 5x, t^2 - 4*t - 12 = 0; D/4 = 2^2 +12 = 4+12 = 16 = 4^2, t1 = (2-4)/1 = -2; t2 = (2+4)/1 = 6. 1) x^2 + 5x = -2; x^2 + 5x + 2 = 0; D1 = 5^2 - 4*2 = 25 - 8 = 17, x1 = (-5 - sqrt(17))/2; x2 = (-5 + sqrt(17))/2; 2) x^2 + 5x = 6; x^2 + 5x - 6 = 0; D2 = 5^2 + 4*6 = 25 + 24 = 49 =7^2; x3 = (-5-7)/2 = -12/2 = -6; x4 = (-5+7)/2 = 2/2 = 1;