0=-4.905t^2+20t-15

Solution for 0=-4.905t^2+20t-15 equation:

0=-4.905t^2+20t-15

We move all terms to the left:

0-(-4.905t^2+20t-15)=0

We add all the numbers together, and all the variables

-(-4.905t^2+20t-15)=0

We get rid of parentheses

4.905t^2-20t+15=0

a = 4.905; b = -20; c = +15;
Δ = b2-4ac
Δ = -202-4·4.905·15
Δ = 105.7
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{-(-20)-\sqrt{105.7}}{2*4.905}=\frac{20-\sqrt{105.7}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+\sqrt{105.7}}{2*4.905}=\frac{20+\sqrt{105.7}}{9.81}
$