T(n)=n^2+3n

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

(T)=T^2+3T

We move all terms to the left:

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

We get rid of parentheses

-T^2+T-3T=0

We add all the numbers together, and all the variables

-1T^2-2T=0

a = -1; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-1)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-1}=\frac{0}{-2}
=0
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-1}=\frac{4}{-2}
=-2
$