H=-4.9t^2+1.7t+4

Solution for H=-4.9t^2+1.7t+4 equation:

=-4.9H^2+1.7H+4

We move all terms to the left:

-(-4.9H^2+1.7H+4)=0

We get rid of parentheses

4.9H^2-1.7H-4=0

a = 4.9; b = -1.7; c = -4;
Δ = b2-4ac
Δ = -1.72-4·4.9·(-4)
Δ = 81.29
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{-(-1.7)-\sqrt{81.29}}{2*4.9}=\frac{1.7-\sqrt{81.29}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.7)+\sqrt{81.29}}{2*4.9}=\frac{1.7+\sqrt{81.29}}{9.8}
$