20=-4.9t^2+23t+1

Solution for 20=-4.9t^2+23t+1 equation:

20=-4.9t^2+23t+1

We move all terms to the left:

20-(-4.9t^2+23t+1)=0

We get rid of parentheses

4.9t^2-23t-1+20=0

We add all the numbers together, and all the variables

4.9t^2-23t+19=0

a = 4.9; b = -23; c = +19;
Δ = b2-4ac
Δ = -232-4·4.9·19
Δ = 156.6
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{-(-23)-\sqrt{156.6}}{2*4.9}=\frac{23-\sqrt{156.6}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{156.6}}{2*4.9}=\frac{23+\sqrt{156.6}}{9.8}
$