0=0.8-5t^2

Solution for 0=0.8-5t^2 equation:

0=0.8-5t^2

We move all terms to the left:

0-(0.8-5t^2)=0

We add all the numbers together, and all the variables

-(0.8-5t^2)=0

We get rid of parentheses

5t^2-0.8=0

a = 5; b = 0; c = -0.8;
Δ = b2-4ac
Δ = 02-4·5·(-0.8)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*5}=\frac{-4}{10}
=-2/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*5}=\frac{4}{10}
=2/5
$