64y^2+)7=32

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Solution for 64y^2+)7=32 equation: 



64y^2+)7=32

We move all terms to the left:

64y^2+)7-(32)=0

We add all the numbers together, and all the variables

64y^2=0

a = 64; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·64·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$y=\frac{-b}{2a}=\frac{0}{128}=0$

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