64=64+48t+16t^2

Solution for 64=64+48t+16t^2 equation:

64=64+48t+16t^2

We move all terms to the left:

64-(64+48t+16t^2)=0

We get rid of parentheses

-16t^2-48t-64+64=0

We add all the numbers together, and all the variables

-16t^2-48t=0

a = -16; b = -48; c = 0;
Δ = b2-4ac
Δ = -482-4·(-16)·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-48}{2*-16}=\frac{0}{-32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+48}{2*-16}=\frac{96}{-32}
=-3
$