If you have the sqrt(3+x) -1 = sqrt(3x+10), how would you solve for x

Please give step by step process. I'm studying for the finals and I found this problem in the book and I can't solve it. HELP!

Isolate the radical with the most complex looking radicand. Because 3x + 10 looks more "complex" than x + 3, we isolate this term. Once that's done, square both sides to get rid of the radical.

sqrt(3x + 10) = sqrt(x + 3) - 1
3x + 10 = [sqrt(x + 3) - 1]^2 = (x + 3) - 2 sqrt(x + 3) + 1 [Note: (a - b)^2 = a^2 - 2ab + b^2]
3x + 10 = x + 4 - 2 sqrt(x + 3)

Isolate the radical once more and repeat the procedure.

-2 sqrt(x + 3) = 2x + 6 iff
sqrt(x + 3) = -x - 3 iff
x + 3 = (-x - 3)^2 = x^2 + 6x + 9 iff
x^2 + 5x + 6 = 0 iff
(x + 2)(x + 3) = 0 iff
x + 2 = 0 or x + 3 = 0 iff
x = -2 or x = -3

You must check your solution for extraneous solution.

If x = -2, the equation is no longer true.
If x = -3, the equation is also false.

Thus there are no solution to this equation.

Square both sides

(sqrt(3 + x) - 1)^2 = (sqrt(3x + 10))^2
(sqrt(3 + x) - 1) * (sqrt(3 + x) - 1) = sqrt(3x + 10) * sqrt(3x + 10)
3 + x - 2 * sqrt(3 + x) + 1 = 3x + 10
4 + x - 2 * sqrt(3 + x) = 3x + 10
-2 * sqrt(3 + x) = 3x - x + 10 - 4
-2 * sqrt(3 + x) = 2x + 6
-sqrt(3 + x) = x + 3

We can square both sides again at this point, or use a substitution. Let's just square both sides and work from there

3 + x = x^2 + 6x + 9
x^2 + 6x - x + 9 - 3 = 0
x^2 + 5x + 6 = 0

Either factor or use the quadratic formula to solve for x

x = (-5 +/- sqrt(5^2 - 4 * 1 * 6)) / (2 * 1)
x = (-5 +/- sqrt(25 - 24)) / 2
x = (-5 +/- sqrt(1)) / 2
x = (-5 +/- 1) / 2
x = -6/2 , -4/2
x = -3 , -2

Now, because we squared (and squared again) both sides of the equation, we may have an extraneous answer (because the square of a negative number is identical to the square of a positive number, i.e. (-4)^2 = 16 = (+4)^2). We'll need to test our answers in the original equation and see how it fares