H(t)=-4.9t^2+24t+15

Solution for H(t)=-4.9t^2+24t+15 equation:

(H)=-4.9H^2+24H+15

We move all terms to the left:

(H)-(-4.9H^2+24H+15)=0

We get rid of parentheses

4.9H^2-24H+H-15=0

We add all the numbers together, and all the variables

4.9H^2-23H-15=0

a = 4.9; b = -23; c = -15;
Δ = b2-4ac
Δ = -232-4·4.9·(-15)
Δ = 823
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{-(-23)-\sqrt{823}}{2*4.9}=\frac{23-\sqrt{823}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{823}}{2*4.9}=\frac{23+\sqrt{823}}{9.8}
$