H(t)=-4.9t^2+26.95t+4

Solution for H(t)=-4.9t^2+26.95t+4 equation:

(H)=-4.9H^2+26.95H+4

We move all terms to the left:

(H)-(-4.9H^2+26.95H+4)=0

We get rid of parentheses

4.9H^2-26.95H+H-4=0

We add all the numbers together, and all the variables

4.9H^2-25.95H-4=0

a = 4.9; b = -25.95; c = -4;
Δ = b2-4ac
Δ = -25.952-4·4.9·(-4)
Δ = 751.8025
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{-(-25.95)-\sqrt{751.8025}}{2*4.9}=\frac{25.95-\sqrt{751.8025}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25.95)+\sqrt{751.8025}}{2*4.9}=\frac{25.95+\sqrt{751.8025}}{9.8}
$