16=-4.9t^2+18t+1

Solution for 16=-4.9t^2+18t+1 equation:

16=-4.9t^2+18t+1

We move all terms to the left:

16-(-4.9t^2+18t+1)=0

We get rid of parentheses

4.9t^2-18t-1+16=0

We add all the numbers together, and all the variables

4.9t^2-18t+15=0

a = 4.9; b = -18; c = +15;
Δ = b2-4ac
Δ = -182-4·4.9·15
Δ = 30
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{-(-18)-\sqrt{30}}{2*4.9}=\frac{18-\sqrt{30}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+\sqrt{30}}{2*4.9}=\frac{18+\sqrt{30}}{9.8}
$