10=80(t)+16t^2

Solution for 10=80(t)+16t^2 equation:

10=80(t)+16t^2

We move all terms to the left:

10-(80(t)+16t^2)=0

We get rid of parentheses

-16t^2-80t+10=0

a = -16; b = -80; c = +10;
Δ = b2-4ac
Δ = -802-4·(-16)·10
Δ = 7040
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:

$\sqrt{\Delta}=\sqrt{7040}=\sqrt{64*110}=\sqrt{64}*\sqrt{110}=8\sqrt{110}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-8\sqrt{110}}{2*-16}=\frac{80-8\sqrt{110}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+8\sqrt{110}}{2*-16}=\frac{80+8\sqrt{110}}{-32}
$