How do you solve the equation #x^2-3x-18=0# by using the quadratic formula?

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Answer 1 · 2018-05-11T11:18:53

#x=-3 and x=6#

It is really worth trying to commit this formula to memory.

Given: #x^2-3x-18=0#

Compare to #ax^2+bx+c=0# where #x=(-b+-sqrt(b^2-4ac))/(2a)#

Comparing the two we have:

#a=1; b=-3 and c=-18#

#x=(-(-3)+-sqrt((-3)^2-4(1)(-18)))/(2(1))#

#x=(3+-sqrt(81))/(2(1))#

#x=(3+-9)/2#

#x=-3 and x=6#