First, we have to move everything to the left so it resembles this form: #ax^2+bx+c=0#, also known as the standard form.

For this case, all we have to do is move everything to 1 side (it doesn't matter which side, although I prefer to move it all to the side with #x^2# already in it)

#x^2-3x=4x-1#

#x^2-7x+1=0#

Now you will need to know the quadratic formula to solve this problem: #x=(-b+-sqrt(b^2-4ac))/(2a)#

But what are #a#, #b#, and #c# you ask? Well that's why we rearranged the original equation. #a# is the number next to #x^2#, #b# is the number next to #x#, and #c# doesn't have an #x# next to it. Doing this, we find #a=1#, #b=-7#, #c=1#. All we have to do is put into the quadratic formula and simplify:

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

#x=(7+-sqrt(49-4))/(2)#

#x=(7+-sqrt(45))/(2)#

So this gives us two answers:

#x=(7+sqrt(45))/(2) or (7-sqrt(45))/(2)#

Which are the correct answers.