0=-4.9t^2+10t+8

Solution for 0=-4.9t^2+10t+8 equation:

0=-4.9t^2+10t+8

We move all terms to the left:

0-(-4.9t^2+10t+8)=0

We add all the numbers together, and all the variables

-(-4.9t^2+10t+8)=0

We get rid of parentheses

4.9t^2-10t-8=0

a = 4.9; b = -10; c = -8;
Δ = b2-4ac
Δ = -102-4·4.9·(-8)
Δ = 256.8
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{-(-10)-\sqrt{256.8}}{2*4.9}=\frac{10-\sqrt{256.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{256.8}}{2*4.9}=\frac{10+\sqrt{256.8}}{9.8}
$