Answer
$t= -1$ or $ t = -19$
Work Step by Step
Divide both sides of the equation by $3$ so that the square term has a coefficient of one. Then, transfer the constant to the right of the equal sign and proceed with completing the square of the equation by dividing the coefficient of the middle term term by $2$ and squaring it and adding the result on both sides.
$\begin{array}{rl}
4t^2+80 t+ 20 & = -56 \\
t^2+20t+ 5 & = -14 \\
t^2+20t & = -5-14 \\
t^2+20t & = -19 \\
t^2+20t +\left(\frac{20}{2}\right)^2 & =-19+\left(\frac{20}{2}\right)^2 \\
t^2+20t +10^2 & =-19+10^2 \\
(t+10)^2 & = 9 \\
t+10 & = \pm \sqrt{81}
\\ \text { Find the two separate solutions:} \\
t+10=9 & \text { or}\ \ t+10=-9 \\
t=-10+9 & \text { or}\ \ t=-10-9 \\
t= -1 & \text { or}\ \ t = -19 \\
\end{array}$
Check
$$\begin{aligned}
4(-1)^2+80(-1)+ 20 & \stackrel{?}{=} -56 \\
-56& = -56\checkmark \\
4(-19)^2+80(-19)+ 20 & \stackrel{?}{=} -56 \\
-56&=-56\checkmark.
\end{aligned}$$ The solution is $t= -1$ or $ t = -19$.
