H(t)=4+5t-4,5t^2

Solution for H(t)=4+5t-4,5t^2 equation:

(H)=4+5H-4.5H^2

We move all terms to the left:

(H)-(4+5H-4.5H^2)=0

We get rid of parentheses

4.5H^2-5H+H-4=0

We add all the numbers together, and all the variables

4.5H^2-4H-4=0

a = 4.5; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·4.5·(-4)
Δ = 88
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{22}}{2*4.5}=\frac{4-2\sqrt{22}}{9}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{22}}{2*4.5}=\frac{4+2\sqrt{22}}{9}
$