The quadratic formula is #(-b+-sqrt(b^2-4ac))/(2a)#.

In a trinomial written in the form #ax^2+bx+c#,

#a#- the coefficient (number) in front of the #x^2#

#b#- the coefficient in front of the #x#

#c#- the constant (number by itself and has no variable- in this case, has no #x#- attached to it

So, in this problem,

#a= 1# (no coefficient is an invisible #1#)

#b= -3# (don't forget about the negative!)

#c= -4#

Now, we plug all these values into the quadratic formula with the corresponding values:

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

#(3+-sqrt(9-4(-4)))/(2)#

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

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

#(3+-5)/(2)#

Now, you have to solve for BOTH solutions:

One solution: #x=# #(3+5)/2# = #8/2# = #4#

Second solution: #x=# #(3-5)/2# = #-2/2# = #-1#

So, the two solutions are #x= -1# and #x=4#