x^2+0,1x=5,8(10^-5)

Solution for x^2+0,1x=5,8(10^-5) equation:

x^2+0.1x=5.8(10^-5)

We move all terms to the left:

x^2+0.1x-(5.8(10^-5))=0

We add all the numbers together, and all the variables

x^2+0.1x-5-5.8E=0

We add all the numbers together, and all the variables

x^2+0.1x-20.766034605062=0

a = 1; b = 0.1; c = -20.766034605062;
Δ = b2-4ac
Δ = 0.12-4·1·(-20.766034605062)
Δ = 83.074138420248
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{-(0.1)-\sqrt{83.074138420248}}{2*1}=\frac{-0.1-\sqrt{83.074138420248}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.1)+\sqrt{83.074138420248}}{2*1}=\frac{-0.1+\sqrt{83.074138420248}}{2}
$