A^3-4a^2+8=0

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Solution for A^3-4a^2+8=0 equation: 



^3-4A^2+8=0

We add all the numbers together, and all the variables

-4A^2=0

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

                              Stosujemy wzór:
$A=\frac{-b}{2a}=\frac{0}{-8}=0$

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