Chapter 9 - Roots and Radicals - 9.5 - Solving Radical Equations - Problem Set 9.5 - Page 424: 23

$y=\dfrac{21}{4}$

Squaring both sides of the equation and then using the properties of equalities, we obtain: \begin{array}{l}\require{cancel}\left( 2\sqrt{y+1} \right)^2=\left( 5 \right)^2 \\\\ 4(y+1)=25 \\\\ 4y+4=25 \\\\ 4y=25-4 \\\\ 4y=21 \\\\ y=\dfrac{21}{4} .\end{array}