H(t)=40-9.8t^2

Solution for H(t)=40-9.8t^2 equation:

(H)=40-9.8H^2

We move all terms to the left:

(H)-(40-9.8H^2)=0

We get rid of parentheses

9.8H^2+H-40=0

a = 9.8; b = 1; c = -40;
Δ = b2-4ac
Δ = 12-4·9.8·(-40)
Δ = 1569
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)-\sqrt{1569}}{2*9.8}=\frac{-1-\sqrt{1569}}{19.6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1569}}{2*9.8}=\frac{-1+\sqrt{1569}}{19.6}
$