How do you solve the equation #2x^2-3x-9=0#?

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Answer 1 · 2017-03-30T12:40:21

#"answer "x_1=-3/2 or x_2=3#

#"if "ax^2+bx+c=0#

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

#"since your equation is "2x^2-3x-9=0,#

#a=2" , "b=-3" , "c=-9#

#Delta=sqrt((-3)^2-4*2*(-9))#

#Delta=sqrt(9+72)=sqrt(81)#

#Delta =+-9#

#x_("1,2")=(+-b +-Delta)/(2a)#

#x_1=(-b-Delta)/(2a)=(3-9)/(2*2)=-6/4=-3/2#

#x_2=(-b+Delta)/(2a)=(3+9)/(2*2)=12/4=3#