How do you solve #x^2- 7x + 12 = 0# using the quadratic formula?

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Answer 1 · 2018-05-02T16:20:43

#x=4# or #3#

According to quadratic formula if #ax^2+bx+c=0#, then #x=(-b+-sqrt(b^2-4ac))/(2a)#

Here we have #x^2-7x+12=0# i.e. #a=1#, #b=-7# and #c=12#

and #x=(-(-7)+-sqrt((-7)^2-4*1*12))/(2*1)#

= #(7+-sqrt(49-48))/2#

= #(7+-sqrt1)/2#

= #(7+-1)/2#

i.e. #x=4# or #3#