t=-16t^2+12.8t+3.2

Solution for t=-16t^2+12.8t+3.2 equation:

t=-16t^2+12.8t+3.2

We move all terms to the left:

t-(-16t^2+12.8t+3.2)=0

We get rid of parentheses

16t^2-12.8t+t-3.2=0

We add all the numbers together, and all the variables

16t^2-11.8t-3.2=0

a = 16; b = -11.8; c = -3.2;
Δ = b2-4ac
Δ = -11.82-4·16·(-3.2)
Δ = 344.04
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11.8)-\sqrt{344.04}}{2*16}=\frac{11.8-\sqrt{344.04}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11.8)+\sqrt{344.04}}{2*16}=\frac{11.8+\sqrt{344.04}}{32}
$