F(x)=2x^2+20x+51

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Solution for F(x)=2x^2+20x+51 equation: 



(F)=2F^2+20F+51

We move all terms to the left:

(F)-(2F^2+20F+51)=0

We get rid of parentheses

-2F^2+F-20F-51=0

We add all the numbers together, and all the variables

-2F^2-19F-51=0

a = -2; b = -19; c = -51;
Δ = b2-4ac
Δ = -192-4·(-2)·(-51)
Δ = -47
Delta is less than zero, so there is no solution for the equation

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