How do you solve #7x^2 - 3x = 2# using the quadratic formula?

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Answer 1 · 2016-03-27T12:51:04

Solution is #x=(3-sqrt(65))/14# and #x=(3+sqrt(65))/14#

According to quadratic formula, the solution of quadratic equation #ax^2+bx+c=0# is given by

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

As #7x^2-3x=2# can be written in form #ax^2+bx+c=0# as

#7x^2-3x-2=0#, (i.e. #a=7#, #b=-3# and #c=-2#) and its solution will be given by

#x=(-(-3)+-sqrt((-3)^2-4xx7xx(-2)))/(2xx7)# or

#x=(3+-sqrt(9+56))/14=(3+-sqrt(65))/14#

Hence, solution is #x=(3-sqrt(65))/14# and #x=(3+sqrt(65))/14#