(2-t)(1-2t)=0

Solution for (2-t)(1-2t)=0 equation:

(2-t)(1-2t)=0

We add all the numbers together, and all the variables

(-1t+2)(-2t+1)=0

We multiply parentheses ..

(+2t^2-1t-4t+2)=0

We get rid of parentheses

2t^2-1t-4t+2=0

We add all the numbers together, and all the variables

2t^2-5t+2=0

a = 2; b = -5; c = +2;
Δ = b2-4ac
Δ = -52-4·2·2
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-3}{2*2}=\frac{2}{4}
=1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+3}{2*2}=\frac{8}{4}
=2
$