3=t^2-t

Solution for 3=t^2-t equation:

3=t^2-t

We move all terms to the left:

3-(t^2-t)=0

We get rid of parentheses

-t^2+t+3=0

We add all the numbers together, and all the variables

-1t^2+t+3=0

a = -1; b = 1; c = +3;
Δ = b2-4ac
Δ = 12-4·(-1)·3
Δ = 13
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{-(1)-\sqrt{13}}{2*-1}=\frac{-1-\sqrt{13}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{13}}{2*-1}=\frac{-1+\sqrt{13}}{-2}
$