-0.8t^2+2t+8=0

Solution for -0.8t^2+2t+8=0 equation:

-0.8t^2+2t+8=0

a = -0.8; b = 2; c = +8;
Δ = b2-4ac
Δ = 22-4·(-0.8)·8
Δ = 29.6
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{-(2)-\sqrt{29.6}}{2*-0.8}=\frac{-2-\sqrt{29.6}}{-1.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{29.6}}{2*-0.8}=\frac{-2+\sqrt{29.6}}{-1.6}
$