0=t^2-2t

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

0=t^2-2t

We move all terms to the left:

0-(t^2-2t)=0

We add all the numbers together, and all the variables

-(t^2-2t)=0

We get rid of parentheses

-t^2+2t=0

We add all the numbers together, and all the variables

-1t^2+2t=0

a = -1; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-1)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-1}=\frac{-4}{-2}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-1}=\frac{0}{-2}
=0
$