F(x)=x^2+40x+500

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Solution for F(x)=x^2+40x+500 equation: 



(F)=F^2+40F+500

We move all terms to the left:

(F)-(F^2+40F+500)=0

We get rid of parentheses

-F^2+F-40F-500=0

We add all the numbers together, and all the variables

-1F^2-39F-500=0

a = -1; b = -39; c = -500;
Δ = b2-4ac
Δ = -392-4·(-1)·(-500)
Δ = -479
Delta is less than zero, so there is no solution for the equation

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