32x^2-8x=0

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Solution for 32x^2-8x=0 equation: 



32x^2-8x=0

a = 32; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·32·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*32}=\frac{0}{64}
=0
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*32}=\frac{16}{64}
=1/4
$

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