F(x)=4x^2+19

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Solution for F(x)=4x^2+19 equation: 



(F)=4F^2+19

We move all terms to the left:

(F)-(4F^2+19)=0

We get rid of parentheses

-4F^2+F-19=0

a = -4; b = 1; c = -19;
Δ = b2-4ac
Δ = 12-4·(-4)·(-19)
Δ = -303
Delta is less than zero, so there is no solution for the equation

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