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0=3x^2+12x+16
We move all terms to the left:
0-(3x^2+12x+16)=0
We add all the numbers together, and all the variables
-(3x^2+12x+16)=0
We get rid of parentheses
-3x^2-12x-16=0
a = -3; b = -12; c = -16;
Δ = b2-4ac
Δ = -122-4·(-3)·(-16)
Δ = -48
Delta is less than zero, so there is no solution for the equation
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