-0.2x2+0.4x+0.8=0

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Solution for -0.2x2+0.4x+0.8=0 equation: 



-0.2x^2+0.4x+0.8=0

a = -0.2; b = 0.4; c = +0.8;
Δ = b2-4ac
Δ = 0.42-4·(-0.2)·0.8
Δ = 0.8
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{-(0.4)-\sqrt{0.8}}{2*-0.2}=\frac{-0.4-\sqrt{0.8}}{-0.4}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{0.8}}{2*-0.2}=\frac{-0.4+\sqrt{0.8}}{-0.4}
$

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