How do you solve #35x^2+6x-9=0 # using the quadratic formula?

1 Answer
Apr 23, 2016

Answer:

#x_1=-6/10#
#x_2=3/7#

Explanation:

#"if an equation is given as "ax^2+bx+c=0;#
#"the quadratic formula is determined b y "x_"1,2"=(-b±sqrt(b^2-4*a*c))/(2*a)#

#35x^2+6x-9=0#
#a=35" ;"b=6" ;"c=-9#

#Delta=sqrt(b^2-4*a*c)=sqrt(6^2+4*35*9)#
#Delta=sqrt(36+1260)=sqrt1296#
#Delta=36#

#x_1=(-b-Delta)/(2*a)" "x_1=(-6-36)/(2*35)" "x_1=-42/70#

#x_1=-6/10#

#x_2=(-b+Delta)/(2*a)" "x_2=(-6+36)/(2*35)" "x_2=30/70" "x_2=3/7#