(8x)^2-4(x)=0

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Solution for (8x)^2-4(x)=0 equation: 



(8x)^2-4(x)=0

a = 8; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·8·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*8}=\frac{0}{16}
=0
$


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

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