15=20t-4.9t^2

Solution for 15=20t-4.9t^2 equation:

15=20t-4.9t^2

We move all terms to the left:

15-(20t-4.9t^2)=0

We get rid of parentheses

4.9t^2-20t+15=0

a = 4.9; b = -20; c = +15;
Δ = b2-4ac
Δ = -202-4·4.9·15
Δ = 106
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{106}}{2*4.9}=\frac{20-\sqrt{106}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+\sqrt{106}}{2*4.9}=\frac{20+\sqrt{106}}{9.8}
$