We take the square root of both sides of the equation and solve the resulting absolute value equation: \begin{align*} \sqrt{q^2} &= \sqrt{25} \\[3pt] \vert q \vert &= 5 \\[3pt] q &= \pm 5 \end{align*}

First, we isolate the $x^2$ on one side of the equation: \begin{align*} 2x^2 - 32 &= 0 \\[3pt] 2x^2 &= 32 \\[3pt] x^2 &= 16 \end{align*}

Then, we take the square root of both sides of the equation and solve the resulting absolute value equation: \begin{align*} \sqrt{x^2} &= \sqrt{16} \\[3pt] \vert x \vert &= 4 \\[3pt] x &= \pm 4 \end{align*}

First, we isolate the $t^2$ on one side of the equation: \begin{align*} \dfrac{t^2}{4} -\dfrac 3 {2} &= \dfrac{3}{4} \\[5pt] \dfrac{t^2}{4} & = \dfrac{3}{4} + \dfrac 3 {2} \\[5pt] \dfrac{t^2}{4} & = \dfrac{3}{4} + \dfrac 6 {4} \\[5pt] \dfrac{t^2}{4} & = \dfrac{9}{4}\\[5pt] t^2 & = 9 \end{align*}

Then, we take the square root of both sides of the equation and solve the resulting absolute value equation: \begin{align*} \sqrt{t^2} & = \sqrt{9}\\[3pt] |t| &= 3\\[3pt] t &= \pm 3 \end{align*}

First, we isolate the $y^2$ on one side of the equation: \begin{align*} 25y^2 &= 4 \\[5pt] y^2 &= \dfrac{4}{25} \end{align*}

Then, we take the square root of both sides of the equation and solve the resulting absolute value equation: \begin{align*} \sqrt{y^2} &= \sqrt{\dfrac{4}{25}} \\[5pt] \vert y\vert &= \dfrac{\sqrt{4}}{\sqrt{25}} \\[5pt] \vert y \vert &= \dfrac{2}{5} \\[5pt] y &=\bbox[4pt,border: 1px solid lightgray]{ \pm \dfrac{2}{5}} \end{align*}

First, we isolate the $y^2$ on one side of the equation: \begin{align*} 9y^2-16 &= 0\\[3pt] 9y^2 &=16\\[3pt] y^2 &=\dfrac{16}{9} \end{align*}

Then, we take the square root of both sides of the equation and solve the resulting absolute value equation: \begin{align*} \sqrt{y^2} &= \sqrt{\dfrac{16}{9}} \\[5pt] \vert y\vert &= \dfrac{\sqrt{16}}{\sqrt{9}} \\[5pt] \vert y \vert &= \dfrac{4}{3}\\[5pt] y &= \pm \dfrac{4}{3} \end{align*}

First, we isolate the $p^2$ on one side of the equation: \begin{align*} 10p^2 - \dfrac 1 {10} &= 0 \\[5pt] 10p^2 &= \dfrac 1 {10} \\[5pt] p^2 &= \dfrac 1 {100} \end{align*}

Then, we take the square root of both sides of the equation and solve the resulting absolute value equation: \begin{align*} \sqrt{p^2} &= \sqrt{\dfrac 1 {100}} \\[5pt] \vert p\vert &= \dfrac{\sqrt 1}{\sqrt{100}} \\[5pt] \vert p \vert &= \dfrac 1 {10} \\[5pt] p &= \pm \dfrac 1 {10} \end{align*}

First, we isolate the $b^2$ on one side of the equation: \begin{align*} b^2 - 1 &= 7 \\[3pt] b^2 &= 8 \end{align*}

Then, we take the square root of both sides of the equation and solve the resulting absolute value equation: \begin{align*} \sqrt{b^2} &= \sqrt{8} \\[3pt] \vert b \vert &= \sqrt{8} \\[3pt] \vert b \vert &= \sqrt{4 \cdot 2} \\[3pt] \vert b \vert &= \sqrt{4} \cdot \sqrt{2} \\[3pt] \vert b \vert &= 2\sqrt{2} \\[3pt] b &= \pm 2\sqrt{2} \end{align*}

First, we isolate the $z^2$ on one side of the equation: \begin{align*} 53-z^2 &= 3 \\[5pt] -z^2 &= -50 \\[5pt] z^2 & = 50 \end{align*}

Then, we take the square root of both sides of the equation and solve the resulting absolute value equation:

\begin{align*} \sqrt{z^2} &= \sqrt{50} \\[5pt] \vert z \vert &= \sqrt{25\cdot 2} \\[3pt] \vert z \vert &= \sqrt{25}\cdot \sqrt{2}\\[3pt] \vert z \vert &= 5\sqrt{2} \\[3pt] z &= \bbox[4pt,border: 1px solid lightgray]{\pm 5\sqrt{2}} \end{align*}

First, we isolate the $p^2$ on one side of the equation: \begin{align*} p^2+\dfrac{3}{2} &= \dfrac{5}{3} \\[5pt] p^2 &= \dfrac{5}{3} - \dfrac{3}{2} \\[5pt] p^2 &= \dfrac{10}{6} - \dfrac{9}{6} \\[5pt] p^2 &= \dfrac{1}{6} \end{align*}

Then, we take the square root of both sides of the equation and solve the resulting absolute value equation: \begin{align*} \sqrt{p^2} &= \sqrt{\dfrac{1}{6}} \\[5pt] \vert p \vert &= \sqrt{\dfrac{1}{6}} \\[5pt] \vert p \vert &= \dfrac{\sqrt 1}{\sqrt{6}} \\[5pt] \vert p \vert &= \dfrac{1}{\sqrt{6}} \\[5pt] p &= \pm \dfrac1{\sqrt{6}} \end{align*}