(2x^2-1x)=0

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Solution for (2x^2-1x)=0 equation:



(2x^2-1x)=0
We get rid of parentheses
2x^2-1x=0
a = 2; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·2·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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

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