2x^2+32x+128=64x

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Solution for 2x^2+32x+128=64x equation: 



2x^2+32x+128=64x

We move all terms to the left:

2x^2+32x+128-(64x)=0

We add all the numbers together, and all the variables

2x^2-32x+128=0

a = 2; b = -32; c = +128;
Δ = b2-4ac
Δ = -322-4·2·128
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{32}{4}=8$

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