1=4.9t^2+8t

Solution for 1=4.9t^2+8t equation:

1=4.9t^2+8t

We move all terms to the left:

1-(4.9t^2+8t)=0

We get rid of parentheses

-4.9t^2-8t+1=0

a = -4.9; b = -8; c = +1;
Δ = b2-4ac
Δ = -82-4·(-4.9)·1
Δ = 83.6
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{-(-8)-\sqrt{83.6}}{2*-4.9}=\frac{8-\sqrt{83.6}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+\sqrt{83.6}}{2*-4.9}=\frac{8+\sqrt{83.6}}{-9.8}
$