H(t)=-4.9t^2+23

Solution for H(t)=-4.9t^2+23 equation:

(H)=-4.9H^2+23

We move all terms to the left:

(H)-(-4.9H^2+23)=0

We get rid of parentheses

4.9H^2+H-23=0

a = 4.9; b = 1; c = -23;
Δ = b2-4ac
Δ = 12-4·4.9·(-23)
Δ = 451.8
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{-(1)-\sqrt{451.8}}{2*4.9}=\frac{-1-\sqrt{451.8}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{451.8}}{2*4.9}=\frac{-1+\sqrt{451.8}}{9.8}
$