0.5x^2+4x+6=0

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Solution for 0.5x^2+4x+6=0 equation:



0.5x^2+4x+6=0
a = 0.5; b = 4; c = +6;
Δ = b2-4ac
Δ = 42-4·0.5·6
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*0.5}=\frac{-6}{1} =-6 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*0.5}=\frac{-2}{1} =-2 $

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