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;