(-0.8t^2)+(8t)-(10)=0

Solution for (-0.8t^2)+(8t)-(10)=0 equation:

(-0.8t^2)+(8t)-(10)=0

We get rid of parentheses

-0.8t^2+8t-10=0

a = -0.8; b = 8; c = -10;
Δ = b2-4ac
Δ = 82-4·(-0.8)·(-10)
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{2}}{2*-0.8}=\frac{-8-4\sqrt{2}}{-1.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{2}}{2*-0.8}=\frac{-8+4\sqrt{2}}{-1.6}
$