H(t)=-10t^2+40t+1

Solution for H(t)=-10t^2+40t+1 equation:

(H)=-10H^2+40H+1

We move all terms to the left:

(H)-(-10H^2+40H+1)=0

We get rid of parentheses

10H^2-40H+H-1=0

We add all the numbers together, and all the variables

10H^2-39H-1=0

a = 10; b = -39; c = -1;
Δ = b2-4ac
Δ = -392-4·10·(-1)
Δ = 1561
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{-(-39)-\sqrt{1561}}{2*10}=\frac{39-\sqrt{1561}}{20}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{1561}}{2*10}=\frac{39+\sqrt{1561}}{20}
$