16t^2-200t-80=0

Solution for 16t^2-200t-80=0 equation:

16t^2-200t-80=0

a = 16; b = -200; c = -80;
Δ = b2-4ac
Δ = -2002-4·16·(-80)
Δ = 45120
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:

$\sqrt{\Delta}=\sqrt{45120}=\sqrt{64*705}=\sqrt{64}*\sqrt{705}=8\sqrt{705}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-200)-8\sqrt{705}}{2*16}=\frac{200-8\sqrt{705}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-200)+8\sqrt{705}}{2*16}=\frac{200+8\sqrt{705}}{32}
$