-16t^2+80+0.1=0

Solution for -16t^2+80+0.1=0 equation:

-16t^2+80+0.1=0

We add all the numbers together, and all the variables

-16t^2+80.1=0

a = -16; b = 0; c = +80.1;
Δ = b2-4ac
Δ = 02-4·(-16)·80.1
Δ = 5126.4
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{-(0)-\sqrt{5126.4}}{2*-16}=\frac{0-\sqrt{5126.4}}{-32}
=-\frac{\sqrt{}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{5126.4}}{2*-16}=\frac{0+\sqrt{5126.4}}{-32}
=\frac{\sqrt{}}{-32}
$