F(x)=2x^2+5x+20

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Solution for F(x)=2x^2+5x+20 equation: 



(F)=2F^2+5F+20

We move all terms to the left:

(F)-(2F^2+5F+20)=0

We get rid of parentheses

-2F^2+F-5F-20=0

We add all the numbers together, and all the variables

-2F^2-4F-20=0

a = -2; b = -4; c = -20;
Δ = b2-4ac
Δ = -42-4·(-2)·(-20)
Δ = -144
Delta is less than zero, so there is no solution for the equation

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