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

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Answer 1 · 2016-07-30T22:13:49

#{(-1-sqrt41)/10,(-1+sqrt 41)/10}#

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

#y=ax^2+bx+c=0#

#"where ;"#

#a=5" ; "b=1" ; "c=-2#

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

#Delta=sqrt(1^2-4*5*(-2))#

#Delta=sqrt(1+4*5*2)#

#Delta=sqrt41#

#x_(1,2)=(-b±Delta)/(2a)#

#x_1=(-1-sqrt 41)/(2*5)#

#x_1=(-1-sqrt41)/10#

#x_2=(-1+sqrt 41)/(2*5)#

#x_2=(-1+sqrt41)/10#