3=-2t^2+8t+.8

Solution for 3=-2t^2+8t+.8 equation:

3=-2t^2+8t+.8

We move all terms to the left:

3-(-2t^2+8t+.8)=0

We get rid of parentheses

2t^2-8t-.8+3=0

We add all the numbers together, and all the variables

2t^2-8t+2.2=0

a = 2; b = -8; c = +2.2;
Δ = b2-4ac
Δ = -82-4·2·2.2
Δ = 46.4
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{-(-8)-\sqrt{46.4}}{2*2}=\frac{8-\sqrt{46.4}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+\sqrt{46.4}}{2*2}=\frac{8+\sqrt{46.4}}{4}
$