5) 
zpět
substituce: 
y2 – 4y – 5 = 0
(y – 5)(y + 1) = 0
y1 = 5 → 
→ x1
= 25 = 32
y2 = -1 → 
→ x2
= 2-1 = 
5) zpět
substituce:
y2 – 4y – 5 = 0
(y – 5)(y + 1) = 0
y1 = 5 → → x1
= 25 = 32
y2 = -1 → → x2
= 2-1 =