0=8.04t-4.9t^2+1.8

Solution for 0=8.04t-4.9t^2+1.8 equation:

0=8.04t-4.9t^2+1.8

We move all terms to the left:

0-(8.04t-4.9t^2+1.8)=0

We add all the numbers together, and all the variables

-(8.04t-4.9t^2+1.8)=0

We get rid of parentheses

4.9t^2-8.04t-1.8=0

a = 4.9; b = -8.04; c = -1.8;
Δ = b2-4ac
Δ = -8.042-4·4.9·(-1.8)
Δ = 99.9216
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{-(-8.04)-\sqrt{99.9216}}{2*4.9}=\frac{8.04-\sqrt{99.9216}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8.04)+\sqrt{99.9216}}{2*4.9}=\frac{8.04+\sqrt{99.9216}}{9.8}
$