Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 624: 43

$x=3$

$Given,\sqrt (5x+10)=5$ Squaring, $5x+10=25$ $5x+10-10=25-10$ $5x=15$ $x=3$ We need to check by putting the obtained solution in the original equation: $L.H.S=\sqrt (5X3+10)=\sqrt 25=5$ $R.H.S=5$ $L.H.S=R.H.S$,hence it is a valid solution.