1=3.2t^2+12.8t+1

Solution for 1=3.2t^2+12.8t+1 equation:

1=3.2t^2+12.8t+1

We move all terms to the left:

1-(3.2t^2+12.8t+1)=0

We get rid of parentheses

-3.2t^2-12.8t-1+1=0

We add all the numbers together, and all the variables

-3.2t^2-12.8t=0

a = -3.2; b = -12.8; c = 0;
Δ = b2-4ac
Δ = -12.82-4·(-3.2)·0
Δ = 163.84
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{-(-12.8)-\sqrt{163.84}}{2*-3.2}=\frac{12.8-\sqrt{163.84}}{-6.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12.8)+\sqrt{163.84}}{2*-3.2}=\frac{12.8+\sqrt{163.84}}{-6.4}
$