Y=3x^2+2x+12

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Solution for Y=3x^2+2x+12 equation: 



=3Y^2+2Y+12

We move all terms to the left:

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

We get rid of parentheses

-3Y^2-2Y-12=0

a = -3; b = -2; c = -12;
Δ = b2-4ac
Δ = -22-4·(-3)·(-12)
Δ = -140
Delta is less than zero, so there is no solution for the equation

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