$x^2-13x+40$ = 0

D = ${\left(-\frac{13}{2}\right)}^2-40$ = $\frac{169}{4}$ $-$ $40$ = $\frac{169}{4}$$-$ $\frac{160}{4}$ = $\frac{9}{4}$

x_1,2 = $\frac{13}{2}$ ± $\sqrt{\frac{9}{4}}$

x₁ = $\frac{13}{2}$ - $\frac{3}{2}$ = $\frac{10}{2}$ = 5

x₂ = $\frac{13}{2}$ + $\frac{3}{2}$ = $\frac{16}{2}$ = 8

L = { $5$; $8$}

$10+x^2+6x$ = 0

sortieren

$x^2+6x+10$ = 0

D = $3^2-10$ = $9$ $-$ $10$ = $-1$

Da die Diskriminante D < 0 ist, hat die quadratische Gleichung keine Lösunng.

L ={}