А)х²-2х_15=0 Б)х²-2х-15=0
В) х²+20х+15=0
Г)х²-22х-23=0
Д) х²-29х+69=0
Е) х²+22х+21=0
решить задачу по формуле D=b²-4ac ,, x1 =-b+ √D/2a

а) х²-2х_15=0

D=b²-4ac=4+60=64

x1 =-b+ √D/2a=2+√64/2=2+8/2=10/2=5

х2 = -b- √D/2a=2-8/2=-6/2= - 3

В) х²+20х+15=0

D=b²-4ac=400-60=340

x1 =-b+ √D/2a= - 20+ √340/2

х2 = -b- √D/2a= - 20 - √340/2

Г)х²-22х-23=0

D=b²-4ac = 484 + 92=576

x1 =-b+ √D/2a= 22+ √576/2= 22+24/2=46/2=23

x2 =-b- √D/2a= 22-24/2= - 2/2= - 1

Д) х²-29х+69=0

D=b²-4ac = 841 - 276 = 565

x1 =-b+ √D/2a= 29+ √565/2

x2 =-b- √D/2a= 29-√565/2

Е) х²+22х+21=0

D=b²-4ac = 484 - 84 = 400

x1 =-b+ √D/2a= - 22+ √400/2 = - 22 + 20/2= - 2/2 = - 1

x2 =-b- √D/2a= - 22 - 20/2 = - 42/2 = - 21