Solve for X

That's an odd way to solve it, but it's correct. To make things easier, you could square both sides of the equation [itex]\sqrt{x+3} = 6[/itex] and solve it easily.

 

Why did you input 33 as x? Well you get to the solution, but that's not how you should solve irrational equations.

[tex]\sqrt{x + 3}[/tex] = 6

Until here, everything is fine but then you have to square both sides and guarantee that the expression that is under the root is not negative (which now can be excluded because we clearly see that 6 is positive, otherwise if we had a variable there, for example [tex]\sqrt{x + 3}[/tex] = x + 2 can be solved by solving a system - ([tex]\sqrt{x + 3}[/tex])^2 = (x + 2)^2; x + 3 [tex]\geq0[/tex])

So follows:

x + 3 = 36
x = 33


I hope you can understand my explanation.


EDIT: Too late