H(t)=-t^2+6t+10

Solution for H(t)=-t^2+6t+10 equation:

(H)=-H^2+6H+10

We move all terms to the left:

(H)-(-H^2+6H+10)=0

We get rid of parentheses

H^2-6H+H-10=0

We add all the numbers together, and all the variables

H^2-5H-10=0

a = 1; b = -5; c = -10;
Δ = b2-4ac
Δ = -52-4·1·(-10)
Δ = 65
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{-(-5)-\sqrt{65}}{2*1}=\frac{5-\sqrt{65}}{2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{65}}{2*1}=\frac{5+\sqrt{65}}{2}
$