0=2(3x^2+1)

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Solution for 0=2(3x^2+1) equation: 



0=2(3x^2+1)

We move all terms to the left:

0-(2(3x^2+1))=0

We add all the numbers together, and all the variables

-(2(3x^2+1))=0



We calculate terms in parentheses: -(2(3x^2+1)), so:

2(3x^2+1)

We multiply parentheses

6x^2+2

Back to the equation:

-(6x^2+2)



We get rid of parentheses

-6x^2-2=0

a = -6; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·(-6)·(-2)
Δ = -48
Delta is less than zero, so there is no solution for the equation

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