If the triangle is right-angled, then we can find two solutions before continuing to find others. If 2x-2 is not the hypotenuse, then 7 must be, because it's the longer of the two known sides: 49=25+4(x-1)^2, so (x-1)^2=6 and x=1+sqrt(6). If 2x-2 is the hypotenuse, then: 4(x-1)^2=49+25=74, so x-1=sqrt(37/2) and x=1+sqrt(37/2).
Now the general case. Drop a perpendicular from angle ABC on to AC at P. Let's refer to angle BAC as simply A. When A is known, the triangle ABC is determined and x can be calculated. So if we can find x in terms of A we know it can be defined for all values of A.
AP=5cosA and PC=7-5cosA, BP=5sinA, so 4(x-1)^2=25sin^2A+(7-5cosA)^2 (BC^2=BP^2+PC^2, by Pythagoras), so 4x^2-8x+4=25sin^2A+49-70cosA+25cos^2A=74-70cosA. So 4x^2-8x-70(1-cosA)=0 or 2x^2-4x-35(1-cosA)=0. Using the quadratic formula we get x=(1+sqrt((37-35cosA)/2). The negative square root will be disallowed in most cases, because BC=2x-2 has to be strictly positive so x>1. (At x=0, A=0 and ABC ceases to be a triangle and B lies on AC.)