Squaring both sides, we get

3x – 5 = (x-1)^2

3x – 5 = x^2 -2x + 1

x^2 -5x +6 = 0

x^2 -2x -3x +6 = 0

x(x -2) – 3(x -2) =0

(x-3) (x-2) =0

Therefore x = 3 & x = 2

Checking the given equation for x = 3

sqrt (3x – 5) = x-1

now substituting value of x as 3, we get

LHS: sqrt [(3 x 3) – 5] = sqrt ( 9-5) = sqrt (4) = 2

RHS : 3 – 1 = 2

AS LHS = RHS

therefore x = 3 is the solution of the given equation.

Checking the given equation for x = 2

sqrt (3x – 5) = x-1

now substituting value of x as 2, we get

LHS: sqrt [(3 x 2) – 5] = sqrt ( 6-5) = sqrt (1) = 1

RHS : 2 – 1 = 1

AS LHS = RHS

therefore x = 2 is also the solution of the given equation.

coz, 3x-5=x.x+1-2x

x.x-5x+6=0

And then make factors.