-4.90t^2+15.5T+28.0=0

Solution for -4.90t^2+15.5T+28.0=0 equation:

-4.90t^2+15.5+28.0=0

We add all the numbers together, and all the variables

-4.9t^2+43.5=0

a = -4.9; b = 0; c = +43.5;
Δ = b2-4ac
Δ = 02-4·(-4.9)·43.5
Δ = 852.6
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{-(0)-\sqrt{852.6}}{2*-4.9}=\frac{0-\sqrt{852.6}}{-9.8}
=-\frac{\sqrt{}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{852.6}}{2*-4.9}=\frac{0+\sqrt{852.6}}{-9.8}
=\frac{\sqrt{}}{-9.8}
$