27+(3x^2-4x)=14x

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Solution for 27+(3x^2-4x)=14x equation: 



27+(3x^2-4x)=14x

We move all terms to the left:

27+(3x^2-4x)-(14x)=0

We add all the numbers together, and all the variables

-14x+(3x^2-4x)+27=0

We get rid of parentheses

3x^2-14x-4x+27=0

We add all the numbers together, and all the variables

3x^2-18x+27=0

a = 3; b = -18; c = +27;
Δ = b2-4ac
Δ = -182-4·3·27
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{18}{6}=3$

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