h=-4.9t^2+12t+17

Solution for h=-4.9t^2+12t+17 equation:

=-4.9H^2+12H+17

We move all terms to the left:

-(-4.9H^2+12H+17)=0

We get rid of parentheses

4.9H^2-12H-17=0

a = 4.9; b = -12; c = -17;
Δ = b2-4ac
Δ = -122-4·4.9·(-17)
Δ = 477.2
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{-(-12)-\sqrt{477.2}}{2*4.9}=\frac{12-\sqrt{477.2}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+\sqrt{477.2}}{2*4.9}=\frac{12+\sqrt{477.2}}{9.8}
$