2,6x+0,56+x^2=0

Solution for 2,6x+0,56+x^2=0 equation:

2.6x+0.56+x^2=0

a = 1; b = 2.6; c = +0.56;
Δ = b2-4ac
Δ = 2.62-4·1·0.56
Δ = 4.52
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{-(2.6)-\sqrt{4.52}}{2*1}=\frac{-2.6-\sqrt{4.52}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.6)+\sqrt{4.52}}{2*1}=\frac{-2.6+\sqrt{4.52}}{2}
$