H(t)=8-9(5t^2)

Solution for H(t)=8-9(5t^2) equation:

(H)=8-9(5H^2)

We move all terms to the left:

(H)-(8-9(5H^2))=0

determiningTheFunctionDomain
-(8-95H^2)+H=0

We get rid of parentheses

95H^2+H-8=0

a = 95; b = 1; c = -8;
Δ = b2-4ac
Δ = 12-4·95·(-8)
Δ = 3041
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{-(1)-\sqrt{3041}}{2*95}=\frac{-1-\sqrt{3041}}{190}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{3041}}{2*95}=\frac{-1+\sqrt{3041}}{190}
$