2(3x^2+16)=32

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Solution for 2(3x^2+16)=32 equation: 



2(3x^2+16)=32

We move all terms to the left:

2(3x^2+16)-(32)=0

We multiply parentheses

6x^2+32-32=0

We add all the numbers together, and all the variables

6x^2=0

a = 6; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·6·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{12}=0$

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