T(n)=(40-n)(40+2n)

Solution for T(n)=(40-n)(40+2n) equation:

(T)=(40-T)(40+2T)

We move all terms to the left:

(T)-((40-T)(40+2T))=0

We add all the numbers together, and all the variables

T-((-1T+40)(2T+40))=0

We multiply parentheses ..

-((-2T^2-40T+80T+1600))+T=0


We calculate terms in parentheses: -((-2T^2-40T+80T+1600)), so:

(-2T^2-40T+80T+1600)

We get rid of parentheses

-2T^2-40T+80T+1600

We add all the numbers together, and all the variables

-2T^2+40T+1600

Back to the equation:

-(-2T^2+40T+1600)


We get rid of parentheses

2T^2-40T+T-1600=0

We add all the numbers together, and all the variables

2T^2-39T-1600=0

a = 2; b = -39; c = -1600;
Δ = b2-4ac
Δ = -392-4·2·(-1600)
Δ = 14321
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{-(-39)-\sqrt{14321}}{2*2}=\frac{39-\sqrt{14321}}{4}
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{14321}}{2*2}=\frac{39+\sqrt{14321}}{4}
$