-0.8t^2+3.2t+6=0

Solution for -0.8t^2+3.2t+6=0 equation:

-0.8t^2+3.2t+6=0

a = -0.8; b = 3.2; c = +6;
Δ = b2-4ac
Δ = 3.22-4·(-0.8)·6
Δ = 29.44
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{-(3.2)-\sqrt{29.44}}{2*-0.8}=\frac{-3.2-\sqrt{29.44}}{-1.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.2)+\sqrt{29.44}}{2*-0.8}=\frac{-3.2+\sqrt{29.44}}{-1.6}
$