2.6-4.1x+x^2=0

Solution for 2.6-4.1x+x^2=0 equation:

2.6-4.1x+x^2=0

a = 1; b = -4.1; c = +2.6;
Δ = b2-4ac
Δ = -4.12-4·1·2.6
Δ = 6.41
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{6.41}}{2*1}=\frac{4.1-\sqrt{6.41}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4.1)+\sqrt{6.41}}{2*1}=\frac{4.1+\sqrt{6.41}}{2}
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