H=-490t^2+75t+12

Solution for H=-490t^2+75t+12 equation:

=-490H^2+75H+12

We move all terms to the left:

-(-490H^2+75H+12)=0

We get rid of parentheses

490H^2-75H-12=0

a = 490; b = -75; c = -12;
Δ = b2-4ac
Δ = -752-4·490·(-12)
Δ = 29145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-75)-\sqrt{29145}}{2*490}=\frac{75-\sqrt{29145}}{980}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-75)+\sqrt{29145}}{2*490}=\frac{75+\sqrt{29145}}{980}
$