H(t)=t2-2t+1

Solution for H(t)=t2-2t+1 equation:

(H)=H2-2H+1

We move all terms to the left:

(H)-(H2-2H+1)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1H^2+3H-1=0

a = -1; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·(-1)·(-1)
Δ = 5
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{-(3)-\sqrt{5}}{2*-1}=\frac{-3-\sqrt{5}}{-2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{5}}{2*-1}=\frac{-3+\sqrt{5}}{-2}
$