-t^2-2t+1=0

Solution for -t^2-2t+1=0 equation:

-t^2-2t+1=0

We add all the numbers together, and all the variables

-1t^2-2t+1=0

a = -1; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·(-1)·1
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{2}}{2*-1}=\frac{2-2\sqrt{2}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{2}}{2*-1}=\frac{2+2\sqrt{2}}{-2}
$