0.8t^2+8t-10=0

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

0.8t^2+8t-10=0

a = 0.8; b = 8; c = -10;
Δ = b2-4ac
Δ = 82-4·0.8·(-10)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{6}}{2*0.8}=\frac{-8-4\sqrt{6}}{1.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{6}}{2*0.8}=\frac{-8+4\sqrt{6}}{1.6}
$