x^2+4.1x+4=0

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Solution for x^2+4.1x+4=0 equation:



x^2+4.1x+4=0
a = 1; b = 4.1; c = +4;
Δ = b2-4ac
Δ = 4.12-4·1·4
Δ = 0.81
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{-(4.1)-\sqrt{0.81}}{2*1}=\frac{-4.1-\sqrt{0.81}}{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.1)+\sqrt{0.81}}{2*1}=\frac{-4.1+\sqrt{0.81}}{2} $

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