How do you solve #2x^2 + 12x= 10 # using the quadratic formula?

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Answer 1 · 2017-05-09T16:14:24

#x=-3+-sqrt14#

first rearrange to the form#" "ax^2+bx+c=0#

we have

#2x^2+12x-10=0#

divide out any common factors to make it simpler to use.

#(2x^2+12x-10=0)-:2#

#=>x^2+6x-5=0#

the formula is

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#a=1, b=6, c=-5#

#x=(-6+-sqrt(6^2-4xx1xx(-5)))/(2xx1)#

#x=(-6+-sqrt(36+20))/2#

#x=(-6+-sqrt56)/2#

#x=-6/2+-sqrt56/2#

#x=-6/2+-2sqrt14/2#

#:.x=-3+-sqrt14#