F(x)=3x^2+x+12

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Solution for F(x)=3x^2+x+12 equation: 



(F)=3F^2+F+12

We move all terms to the left:

(F)-(3F^2+F+12)=0

We get rid of parentheses

-3F^2+F-F-12=0

We add all the numbers together, and all the variables

-3F^2-12=0

a = -3; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·(-3)·(-12)
Δ = -144
Delta is less than zero, so there is no solution for the equation

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