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

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

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

We move all terms to the left:

t(1-(2t))=0

We add all the numbers together, and all the variables

t(-2t+1)=0

We multiply parentheses

-2t^2+t=0

a = -2; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-2)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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