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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2H^2-39H-10=0

a = 2; b = -39; c = -10;
Δ = b2-4ac
Δ = -392-4·2·(-10)
Δ = 1601
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{1601}}{2*2}=\frac{39-\sqrt{1601}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{1601}}{2*2}=\frac{39+\sqrt{1601}}{4}
$