Question 1088968
First of all, we need to set the restriction to allow the expression to be well defined.

4x - 3 >=0 and 2x - 5 >=0

 x >= 3/4 and x>=5/2

Restriction: x>=5/2

now we solve by squaring both sides:

4x-3 = 4 +4sqrt(2x - 5) + (2x-5) 

2x-2 = 4sqrt(2x-5) = 2(x-1)

(x-1) is positive according to the restriction, so we can squaring both sides

4(x^2-2x+1) = 16(2x-5)

x^2-2x+1 = 8x-20

x^2-10x+21 = 0

(x-3)(x-7) = 0

so acoording to the last expression we have 2 solution x={3,7}

but according to the restriction x>=5/2 so we don't have to discard any solution.

Final Solution={3,7}



@natolino_