Answer
$t=\left\{ \dfrac{1-2\sqrt{6}}{4},\dfrac{1+2\sqrt{6}}{4} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
(1-4p)^2=24
,$ take the square root of both sides (Square Root Property) and simplify the radical. Then use the properties of equality to isolate the variable.
$\bf{\text{Solution Details:}}$
Taking the square root of both sides (Square Root Property), the equation above is equivalent to
\begin{array}{l}\require{cancel}
1-4p=\pm\sqrt{24}
.\end{array}
Simplifying the radical and then using the properties of equality to isolate the variable, the equation above is equivalent to
\begin{array}{l}\require{cancel}
1-4p=\pm\sqrt{4\cdot6}
\\\\
1-4p=\pm\sqrt{(2)^2\cdot6}
\\\\
1-4p=\pm2\sqrt{6}
\\\\
-4p=-1\pm2\sqrt{6}
\\\\
p=\dfrac{-1\pm2\sqrt{6}}{-4}
\\\\
p=\dfrac{1\pm2\sqrt{6}}{4}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
p=\dfrac{1-2\sqrt{6}}{4}
\\\\\text{OR}\\\\
p=\dfrac{1+2\sqrt{6}}{4}
.\end{array}
Hence, $
t=\left\{ \dfrac{1-2\sqrt{6}}{4},\dfrac{1+2\sqrt{6}}{4} \right\}
.$
