32x^2-48x+16=0

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Solution for 32x^2-48x+16=0 equation:



32x^2-48x+16=0
a = 32; b = -48; c = +16;
Δ = b2-4ac
Δ = -482-4·32·16
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16}{2*32}=\frac{32}{64} =1/2 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16}{2*32}=\frac{64}{64} =1 $

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