Chapter 6 - Radical Functions and Rational Exponents - 6-5 Solving Square Root and Other Radical Equations - Practice and Problem-Solving Exercises - Page 395: 16

$23$

Add $7$ to both sides: $\sqrt {2x+3}-7+7=0+7$ $\sqrt {2x+3}=7$ Square both sides: $(\sqrt {2x+3})^2=7^2$ $2x+3=49$ Subtract $3$ from both sides: $2x+3-3=49-3$ $2x=46$ Divide both sides by $2$: $\dfrac{2x}{2}=\dfrac{46}{2}$ $x=23$ Substitute into original problem: $\sqrt {2(23)+3}-7=0$ $\sqrt {46+3}-7=0$ $\sqrt {49}-7=0$ $7-7=0$ $0=0$