How do you solve #x^2-5x-1 = 0# using the quadratic formula?

1 Answer
Apr 14, 2016

Answer:

#x_1=0,195#
#x_2=5,195#

Explanation:

#ax^2+bx+c=0#

#x_"1,2"=(-b±sqrt(b^2-4*a*c))/(2*a)" the quadratic formula"#

#x^2-5x-1=0#

#a=1" "b=-5" "c=-1#

#Delta=sqrt(b^2-4*a*c)#

#Delta=sqrt(25+4*1*1)#

#Delta=±5,39#

#x_1=(-b-Delta)/(2*a)#

#x_1=(5-5,39)/(2*1)#

#x_1=(0,39)/2#

#x_1=0,195#

#x_2=(-b+Delta)/(2*a)#

#x_2=(5+5,39)/2#

#x_2=(10,39)/2#

#x_2=5,195#