0=-16t^2+80t+3.5

Solution for 0=-16t^2+80t+3.5 equation:

0=-16t^2+80t+3.5

We move all terms to the left:

0-(-16t^2+80t+3.5)=0

We add all the numbers together, and all the variables

-(-16t^2+80t+3.5)=0

We get rid of parentheses

16t^2-80t-3.5=0

a = 16; b = -80; c = -3.5;
Δ = b2-4ac
Δ = -802-4·16·(-3.5)
Δ = 6624
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:

$\sqrt{\Delta}=\sqrt{6624}=\sqrt{144*46}=\sqrt{144}*\sqrt{46}=12\sqrt{46}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-12\sqrt{46}}{2*16}=\frac{80-12\sqrt{46}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+12\sqrt{46}}{2*16}=\frac{80+12\sqrt{46}}{32}
$