H=-490t^2+1260t

Solution for H=-490t^2+1260t equation:

=-490H^2+1260H

We move all terms to the left:

-(-490H^2+1260H)=0

We get rid of parentheses

490H^2-1260H=0

a = 490; b = -1260; c = 0;
Δ = b2-4ac
Δ = -12602-4·490·0
Δ = 1587600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1587600}=1260$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1260)-1260}{2*490}=\frac{0}{980}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1260)+1260}{2*490}=\frac{2520}{980}
=2+4/7
$