H(t)=-5t^2+22t+1.2

Solution for H(t)=-5t^2+22t+1.2 equation:

(H)=-5H^2+22H+1.2

We move all terms to the left:

(H)-(-5H^2+22H+1.2)=0

We get rid of parentheses

5H^2-22H+H-1.2=0

We add all the numbers together, and all the variables

5H^2-21H-1.2=0

a = 5; b = -21; c = -1.2;
Δ = b2-4ac
Δ = -212-4·5·(-1.2)
Δ = 465
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{-(-21)-\sqrt{465}}{2*5}=\frac{21-\sqrt{465}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{465}}{2*5}=\frac{21+\sqrt{465}}{10}
$