T(n)=n^2-3

Solution for T(n)=n^2-3 equation:

(T)=T^2-3

We move all terms to the left:

(T)-(T^2-3)=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}
$