T(n)=2n^2-3

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

(T)=2T^2-3

We move all terms to the left:

(T)-(2T^2-3)=0

We get rid of parentheses

-2T^2+T+3=0

a = -2; b = 1; c = +3;
Δ = b2-4ac
Δ = 12-4·(-2)·3
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*-2}=\frac{-6}{-4}
=1+1/2
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*-2}=\frac{4}{-4}
=-1
$