3(1-t2)-2(1-t)+10=(1-t)

Solution for 3(1-t2)-2(1-t)+10=(1-t) equation:

3(1-t2)-2(1-t)+10=(1-t)

We move all terms to the left:

3(1-t2)-2(1-t)+10-((1-t))=0

We add all the numbers together, and all the variables

3(-1t^2+1)-2(-1t+1)-((-1t+1))+10=0

We multiply parentheses

-3t^2+2t-((-1t+1))+3-2+10=0


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

(-1t+1)

We get rid of parentheses

-1t+1

Back to the equation:

-(-1t+1)


We add all the numbers together, and all the variables

-3t^2+2t-(-1t+1)+11=0

We get rid of parentheses

-3t^2+2t+1t-1+11=0

We add all the numbers together, and all the variables

-3t^2+3t+10=0

a = -3; b = 3; c = +10;
Δ = b2-4ac
Δ = 32-4·(-3)·10
Δ = 129
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{-(3)-\sqrt{129}}{2*-3}=\frac{-3-\sqrt{129}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{129}}{2*-3}=\frac{-3+\sqrt{129}}{-6}
$