How do you solve #x^2 - 3x - 4 = 0# using the quadratic formula?

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Answer 1 · 2015-07-22T08:57:55

Identify the coefficients #a#, #b# and #c# and substitute those values into the quadratic formula to find #x=-1# or #x=4#.

#x^2-3x-4 = 0# is of the form #ax^2+bx+c = 0#, with #a=1#, #b=-3# and #c = -4#.

The quadratic formula gives us solutions:

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

#=(3+-sqrt(3^2-(4xx1xx-4)))/(2*1)#

#=(3+-sqrt(9+16))/2#

#=(3+-sqrt(25))/2#

#=(3+-sqrt(5^2))/2#

#=(3+-5)/2#

That is #x=-1# or #x=4#