-16.4t^2+80t-70=0

Solution for -16.4t^2+80t-70=0 equation:

-16.4t^2+80t-70=0

a = -16.4; b = 80; c = -70;
Δ = b2-4ac
Δ = 802-4·(-16.4)·(-70)
Δ = 1808
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:

$\sqrt{\Delta}=\sqrt{1808}=\sqrt{16*113}=\sqrt{16}*\sqrt{113}=4\sqrt{113}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-4\sqrt{113}}{2*-16.4}=\frac{-80-4\sqrt{113}}{-32.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+4\sqrt{113}}{2*-16.4}=\frac{-80+4\sqrt{113}}{-32.8}
$