H(t)=-t^2+2t+26

Solution for H(t)=-t^2+2t+26 equation:

(H)=-H^2+2H+26

We move all terms to the left:

(H)-(-H^2+2H+26)=0

We get rid of parentheses

H^2-2H+H-26=0

We add all the numbers together, and all the variables

H^2-1H-26=0

a = 1; b = -1; c = -26;
Δ = b2-4ac
Δ = -12-4·1·(-26)
Δ = 105
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{105}}{2*1}=\frac{1-\sqrt{105}}{2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{105}}{2*1}=\frac{1+\sqrt{105}}{2}
$