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b) -2/5(2t-5)²=0
wzorek (a+b)²=a²+2ab+b²
-2/5(4t²+20t+25)=0 /*5
-2(4t²+20t+25)=0
-8t²-40t-50=0
delta=b²-4ac= (-40)²-4*(-8)*(-50)=1600-1600=0
t1/2= -b+ - pierwz delty / 2a = -40/-16 i 40/-16 t1=2,5 t2=-2,5
c) √3b²=√27b
√3b² - √27b=0
delta= 27-4*√3*0= delt √27 = 3√3
b1/2= -√27+-3√3 /2*√3
b1=0
b2= -3√3
Mam nadzieje ,że dobrze