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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(2(t-t^2)), so:

2(t-t^2)

We multiply parentheses

-2t^2+2t

Back to the equation:

-(-2t^2+2t)


We get rid of parentheses

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

We add all the numbers together, and all the variables

4t^2-7t+2=0

a = 4; b = -7; c = +2;
Δ = b2-4ac
Δ = -72-4·4·2
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{17}}{2*4}=\frac{7-\sqrt{17}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{17}}{2*4}=\frac{7+\sqrt{17}}{8}
$