X2+0.1x=10^-5

Solution for X2+0.1x=10^-5 equation:

X2+0.1X=10^-5

We move all terms to the left:

X2+0.1X-(10^-5)=0

We add all the numbers together, and all the variables

X^2+0.1X-5-1.0E=0

We add all the numbers together, and all the variables

X^2+0.1X-7.718281828459=0

a = 1; b = 0.1; c = -7.718281828459;
Δ = b2-4ac
Δ = 0.12-4·1·(-7.718281828459)
Δ = 30.883127313836
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{30.883127313836}}{2*1}=\frac{-0.1-\sqrt{30.883127313836}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.1)+\sqrt{30.883127313836}}{2*1}=\frac{-0.1+\sqrt{30.883127313836}}{2}
$