3a2+14a2=(3+)a2=

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Solution for 3a2+14a2=(3+)a2= equation: 



3a^2+14a^2=(3+)a2=

We move all terms to the left:

3a^2+14a^2-((3+)a2)=0

We add all the numbers together, and all the variables

3a^2+14a^2-(0a2)=0

We add all the numbers together, and all the variables

16a^2=0

a = 16; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·16·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$a=\frac{-b}{2a}=\frac{0}{32}=0$

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