0.2x^2-4x+8=0

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Solution for 0.2x^2-4x+8=0 equation:



0.2x^2-4x+8=0
a = 0.2; b = -4; c = +8;
Δ = b2-4ac
Δ = -42-4·0.2·8
Δ = 9.6
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{-(-4)-\sqrt{9.6}}{2*0.2}=\frac{4-\sqrt{9.6}}{0.4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{9.6}}{2*0.2}=\frac{4+\sqrt{9.6}}{0.4} $

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