H=4.9t^2+16t+2

Solution for H=4.9t^2+16t+2 equation:

=4.9H^2+16H+2

We move all terms to the left:

-(4.9H^2+16H+2)=0

We get rid of parentheses

-4.9H^2-16H-2=0

a = -4.9; b = -16; c = -2;
Δ = b2-4ac
Δ = -162-4·(-4.9)·(-2)
Δ = 216.8
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{-(-16)-\sqrt{216.8}}{2*-4.9}=\frac{16-\sqrt{216.8}}{-9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+\sqrt{216.8}}{2*-4.9}=\frac{16+\sqrt{216.8}}{-9.8}
$