Need badly
what is the solution

Sqrt(1-3x)=x+3
[sqrt(1-3x)]^2=(x+3)^2
1-3x=(x+3)(x+3)
1-3x=x^2+6x+9
-3x=x^2+6x+8
0=x^2+9x+8

The answers are -1 and -8 BUT, we have to plug them back into the original equation to make sure we don't get a negative under the square root sign.

After doing this, we realize that only -1 works, so the answer is x=-1

Sorry that this took forever to answer. I was thinking of a good way to explain this, and if you need any further explanation, message me:)

Best wishes:)