H=-4.9t^2+10t+1.5

Solution for H=-4.9t^2+10t+1.5 equation:

=-4.9H^2+10H+1.5

We move all terms to the left:

-(-4.9H^2+10H+1.5)=0

We get rid of parentheses

4.9H^2-10H-1.5=0

a = 4.9; b = -10; c = -1.5;
Δ = b2-4ac
Δ = -102-4·4.9·(-1.5)
Δ = 129.4
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{-(-10)-\sqrt{129.4}}{2*4.9}=\frac{10-\sqrt{129.4}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{129.4}}{2*4.9}=\frac{10+\sqrt{129.4}}{9.8}
$