Answer
$t=\dfrac{3}{4}$
Work Step by Step
Squaring both sides of the given equation, $
4t=\sqrt{8t+3}
,$ results to
\begin{align*}\require{cancel}
(4t)^2&=\left(\sqrt{8t+3}\right)^2
\\
16t^2&=8t+3
\\
16t^2-8t-3&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(4t-3)(4t+1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
4t-3=0 & 4t+1=0
\\
4t=3 & 4t=-1
\\\\
t=\dfrac{3}{4} & t=-\dfrac{1}{4}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }t=\dfrac{3}{4}: & \text{If }t=-\dfrac{1}{4}:
\\\\
4\left(\dfrac{3}{4}\right)\overset{?}=\sqrt{8\left(\dfrac{3}{4}\right)+3} &
4\left(-\dfrac{1}{4}\right)\overset{?}=\sqrt{8\left(-\dfrac{1}{4}\right)+3}
\\\\
\cancelto14\left(\dfrac{3}{\cancelto14}\right)\overset{?}=\sqrt{\cancelto28\left(\dfrac{3}{\cancelto14}\right)+3} &
\cancelto14\left(-\dfrac{1}{\cancelto14}\right)\overset{?}=\sqrt{\cancelto28\left(-\dfrac{1}{\cancelto14}\right)+3}
\\\\
3\overset{?}=\sqrt{6+3} &
-1\overset{?}=\sqrt{-2+3}
\\
3\overset{?}=\sqrt{9} &
-1\overset{?}=\sqrt{1}
\\
3\overset{\checkmark}=3 &
-1\ne1
.\end{array}
Since $t=-\dfrac{1}{4}$ does not satisfy the original equation, then the only solution of the equation $
4t=\sqrt{8t+3}
$ is $t=\dfrac{3}{4}$.
