64x^2+100=160x

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Solution for 64x^2+100=160x equation: 



64x^2+100=160x

We move all terms to the left:

64x^2+100-(160x)=0

a = 64; b = -160; c = +100;
Δ = b2-4ac
Δ = -1602-4·64·100
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{160}{128}=1+1/4$

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