0.5(t)=4.9t^2+80t+0.5

Solution for 0.5(t)=4.9t^2+80t+0.5 equation:

0.5(t)=4.9t^2+80t+0.5

We move all terms to the left:

0.5(t)-(4.9t^2+80t+0.5)=0

We get rid of parentheses

-4.9t^2+0.5t-80t-0.5=0

We add all the numbers together, and all the variables

-4.9t^2-79.5t-0.5=0

a = -4.9; b = -79.5; c = -0.5;
Δ = b2-4ac
Δ = -79.52-4·(-4.9)·(-0.5)
Δ = 6310.45
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{-(-79.5)-\sqrt{6310.45}}{2*-4.9}=\frac{79.5-\sqrt{6310.45}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-79.5)+\sqrt{6310.45}}{2*-4.9}=\frac{79.5+\sqrt{6310.45}}{-9.8}
$