The problem: take two solutions and plug them into the original equation. To check for extraneous solutions.
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Answer: х=3.
Step-by-step explanation:
А point is missing here: determining the allowable values of x for this equation.
1. Consider the right side of the equation:
The root expression 2x+3 must be greater than or equal to 0.
2x+3≥0
2x≥-3\ |:2
x≥-1,5.
2. Consider the left side of the equation:
the left side of the equation must be greater than or equal to 0.
x≥0.
We get x≥0. ⇒
х=3 ∈ х≥0
х=-1 ∉ х≥0.