T=n2+4n+2

Solution for T=n2+4n+2 equation:

=T2+4T+2

We move all terms to the left:

-(T2+4T+2)=0

We add all the numbers together, and all the variables

-(+T^2+4T+2)=0

We get rid of parentheses

-T^2-4T-2=0

We add all the numbers together, and all the variables

-1T^2-4T-2=0

a = -1; b = -4; c = -2;
Δ = b2-4ac
Δ = -42-4·(-1)·(-2)
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{2}}{2*-1}=\frac{4-2\sqrt{2}}{-2}
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{2}}{2*-1}=\frac{4+2\sqrt{2}}{-2}
$