<p><span style="color:#27ae60;"><span style="font-size:20px;">Solving the Problem</span></span></p>
<p>We're given a function and asked to find suitable value $t$, such that putting $2t$ into the function outputs $30$.</p>
<p><strong>Method 1 (Algebra)</strong></p>
<p>First, let's make the problem a bit less intimidating by changing $2t$ to something easier, in this case $a$:</p>
<p>$a = 2t$</p>
<p>So, $f(2t) = f(a) = a^2 + a$</p>
<p>So, now we can solve for a value of a that makes this function equal to $30$:</p>
<ul>
<li>$a^2 + a = 30$</li>
<li>$a^2 + a - 30 = 0$</li>
</ul>
<p>Since we have a quadratic equation here, we just need to solve it with any standard method. This one is pretty straightforward to factorize:</p>
<ul>
<li>$a^2 + a - 30 = 0$</li>
<li>$(a + 6)(a - 5) = 0$</li>
</ul>
<p>So, we know that $a + 6 = 0$ OR $a - 5 = 0$, so, we can deduce that $a = -6$ or $a = 5$.</p>
<p>However, we should remember that $a = 2t$, since we are looking for values of $t$, we have to halve these. So, $t = -3$ or $t = \frac{5}{2}$.</p>
<p><strong>Method 2 (Back Solving)</strong></p>
<p>Since we only have 5 answer choices here, we can also backsolve to find suitable values for t:</p>
<table border="1" cellpadding="1" cellspacing="1" style="width:500px;">
<thead>
<tr>
<th scope="col">$t$</th>
<th scope="col">$2t$</th>
<th scope="col">$f(2t)$</th>
<th scope="col">Is $f(2t)=30$</th>
</tr>
</thead>
<tbody>
<tr>
<td>$-5$</td>
<td>$-10$</td>
<td>$(-10)^2+(-10) = 90$</td>
<td>No</td>
</tr>
<tr>
<td>$-3$</td>
<td>$-6$</td>
<td>$(-6)^2+(-6)=30$</td>
<td>Yes</td>
</tr>
<tr>
<td>$-\frac{1}{2}$</td>
<td>$-1$</td>
<td>$(-1)^2+(-1)=0$</td>
<td>No</td>
</tr>
<tr>
<td>$2$</td>
<td>$4$</td>
<td>$(4)^2+(4)=20$</td>
<td>No</td>
</tr>
<tr>
<td>$\frac{5}{2}$</td>
<td>$5$</td>
<td>$(5)^2+(5)=30$</td>
<td>Yes</td>
</tr>
</tbody>
</table>
<p>As we can see, the only suitable values for $t$ are $-3$ and $\frac{5}{2}$.</p>
<p>So, the only valid answers are <span style="color:#27ae60;">B</span> and <span style="color:#27ae60;">E</span>.</p>