2t-(t-1)-1-2=3t(2-1t)+4

Solution for 2t-(t-1)-1-2=3t(2-1t)+4 equation:

2t-(t-1)-1-2=3t(2-1t)+4

We move all terms to the left:

2t-(t-1)-1-2-(3t(2-1t)+4)=0

We add all the numbers together, and all the variables

2t-(t-1)-(3t(-1t+2)+4)-1-2=0

We add all the numbers together, and all the variables

2t-(t-1)-(3t(-1t+2)+4)-3=0

We get rid of parentheses

2t-t-(3t(-1t+2)+4)+1-3=0


We calculate terms in parentheses: -(3t(-1t+2)+4), so:

3t(-1t+2)+4

We multiply parentheses

-3t^2+6t+4

Back to the equation:

-(-3t^2+6t+4)


We add all the numbers together, and all the variables

-(-3t^2+6t+4)+t-2=0

We get rid of parentheses

3t^2-6t+t-4-2=0

We add all the numbers together, and all the variables

3t^2-5t-6=0

a = 3; b = -5; c = -6;
Δ = b2-4ac
Δ = -52-4·3·(-6)
Δ = 97
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{-(-5)-\sqrt{97}}{2*3}=\frac{5-\sqrt{97}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{97}}{2*3}=\frac{5+\sqrt{97}}{6}
$