H(t)=-4.9t^2+27t+2.4

Solution for H(t)=-4.9t^2+27t+2.4 equation:

(H)=-4.9H^2+27H+2.4

We move all terms to the left:

(H)-(-4.9H^2+27H+2.4)=0

We get rid of parentheses

4.9H^2-27H+H-2.4=0

We add all the numbers together, and all the variables

4.9H^2-26H-2.4=0

a = 4.9; b = -26; c = -2.4;
Δ = b2-4ac
Δ = -262-4·4.9·(-2.4)
Δ = 723.04
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{-(-26)-\sqrt{723.04}}{2*4.9}=\frac{26-\sqrt{723.04}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+\sqrt{723.04}}{2*4.9}=\frac{26+\sqrt{723.04}}{9.8}
$