Respuesta :
t²+ 2t + 1 = 0
a=1, b=2 & c=1 The formula is x'=[-b+√(b²-4ac)]/(2a)
and====================>x"=[-b-√(b²-4ac)]/(2a)
x' = [-2+√(2²-4(1)] / 2(1) ===> x' = -2/2 ==> x' =-1 & x" = -1
x'=x"= -1
a=1, b=2 & c=1 The formula is x'=[-b+√(b²-4ac)]/(2a)
and====================>x"=[-b-√(b²-4ac)]/(2a)
x' = [-2+√(2²-4(1)] / 2(1) ===> x' = -2/2 ==> x' =-1 & x" = -1
x'=x"= -1