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a=6, b=-5 c=1
Δ = b² - 4ac
Δ = (-5)² - 4*6*1
Δ = 25 - 24
Δ > 0 zatem są 2 rozwiązania x1, x2
x1 = (-b-√Δ) / 2a
x1 = (5 -1) / 2*6
x1 = 4/12
x1 = 1/3
x2 = (-b+√Δ) / 2a
x2 = 5+1 /2*6
x2 = 6/12
x2 = 1/2
można sprawdzić np. korzystając ze wzorów Vieta
x1* x2 = c/a
1/2 * 1/3 = 1/6
1/6 = 1/6
L=P
x1 + x2 = -b/a
1/2 + 1/3 = 5/6
3/6 + 2/6 = 5/6
5/6 = 5/6
L=P