Doubt in testing the solution of a quadratic equation So i just stumbled upon something while solving this equation
$\sqrt{6x-2} + 5 - 3x = 0$
$\sqrt{6x-2} = 3x - 5$
Squaring on both sides
$6x-2 = 9x^2 + 25 - 30x$
$9x^2 - 36x + 27 = 0$
$x^2 - 4x + 3 = 0$
$x^2 - 3x - x + 3 = 0$
$x(x - 3) -1( x - 3) = 0$
$(x - 1)( x - 3) = 0$
$x = 1,3$
But when I plug in the values to verify the solution,
$\sqrt{6x-2} + 5 - 3x = 0$
$\sqrt{6(1)-2} + 5 - 3(1) = 0$
$\sqrt{4} + 5 - 3 = 0$
The value of $\sqrt{4}$ is 2 and -2
When I use -2 the equation is satisfied , but when I use +2 the its not
Similarly when I plug in 3
$\sqrt{6(3)-2} + 5 - 3(3) = 0$
$\sqrt{16} + 5 - 9 = 0$
Now if I consider +4 , the equality is satisfied but when I consider -4 it is not.
So what's the correct solution to the problem?

The original equation we are to solve is $ \ \sqrt{6x-2} \ + \ 5 - 3x \ = \ 0 \ \ , $ which can be written (as you did) as the curve intersection equation
$ \ \sqrt{6x-2} \ = \ 3x - 5 \ \ . $  The left side represents a square-root curve, which looks like the "upper half" ( $ \ y \ \ge \ 0 \ $ ) of a "horizontal" parabola with its vertex at $ \ x \ = \ \frac13 \ $ (since the domain of this function is $ \ x  \ \ge  \ \frac13 \ \ . $  The right side represents a "steeply-rising" line with its $ \ y-$ intercept at $ \ (0 \ , \ -5) \ . $ So we would expect that these curves intersect only once.

We can look at what happens with the solutions to this equation in a couple of ways.  When both sides of the equation are "squared" to produce $ \ 6x-2 \ = \ (3x - 5)^2 \ \ , $ the character of the geometric situation is changed:  the square-root curve becomes a straight line and the former line appears as an "upward-opening" parabola.  This new line also has a steep positive slope; since its $ \ y-$intercept, $ \ (0 \ , \ -2) \ $ is "to the left" and only slightly "below" the vertex of the parabola at $ \ \left( \frac53 \ , \ 0 \right) \ \ , $ these two curves will have two intersections which you found from $ \ 9x^2 - 36x + 27 \  = \ 9·(x - 1)·(x - 3) \ = \ 0 \ \ . $
However, the "half" of the parabola "to the left" of the parabola's symmetry axis at $ \ x \ = \ \frac53 \ $ is falsely "generated" by the act of squaring the function $ \ y \ = \ 3x - 5 \ \ . $  Therefore, the solution $ \ x \ = \ 1 \ < \ \frac53 \ \ $ is "spurious" and does not constitute a solution to the original equation (as you found upon inserting  $ \ \sqrt{6·1 \ - \ 2} \ + \ 5 \ - \ 3·1 \ = \ \sqrt4 + 5 - 3 \ \neq \ 0 \ \ ) \ .  $
On a separate question you raised, the single solution $ \ x \ = \ 3 \ $ inserted into this equation yields $$   \sqrt{6·3 \ - \ 2} \ + \ 5 \ - \ 3·3 \ \ = \ \ \sqrt{16}  \ + \  5 \  - \ 9 \ \ = \ \ (+4) \ + \ 5 \ - \ 9 \ \ \overbrace{=}^{!} \ \ 0 \ \   .  $$
The square-root operation only produces a non-negative value, so the appearance of $ \ \sqrt{16} \ $ does not also call for $ \ (-4) \ $ to be used in the equation.

We can also describe this in terms of the intersection(s) of the square-root curve and the line $ \ y \ = \ 3x - 5 \ \ . $  As was discussed earlier, the curve $ \ y \ = \ \sqrt{6x - 2} \ \ $ will only have $ \ y \ \ge \ 0 \ \   $  [the red curve in the graph below], so there can only be the single intersection with the line at $ \ (3 \ , \ 3·3 - 5 = 4) \ \ . $  However, when the original equation is squared, it allows the creation of a second equation which is also consistent with that squared equation, namely, $ \ -\sqrt{6x-2} \ = \ 3x - 5 \ \ . $  This new [orange] curve has $ \ y \ \le \ 0 \ \ $ and has a single intersection with the line $ \ y \ = \ 3x - 5 \   $ at $ \ (1 \ , \ 3·1 - 5 = -2) \ \ . $  Again, this "negative square-root" curve is not implied in the original equation, so we simply disregard this result.  The only solution for our equation is $ \ \mathbf{ x \ = \ 3 \ } \ \ . $
