0.5x^2+0.5x+0.0625=0

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Solution for 0.5x^2+0.5x+0.0625=0 equation:



0.5x^2+0.5x+0.0625=0
a = 0.5; b = 0.5; c = +0.0625;
Δ = b2-4ac
Δ = 0.52-4·0.5·0.0625
Δ = 0.125
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.5)-\sqrt{0.125}}{2*0.5}=\frac{-0.5-\sqrt{0.125}}{1} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.125}}{2*0.5}=\frac{-0.5+\sqrt{0.125}}{1} $

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