x^2+0.1x=0.9975

Solution for x^2+0.1x=0.9975 equation:

x^2+0.1x=0.9975

We move all terms to the left:

x^2+0.1x-(0.9975)=0

We add all the numbers together, and all the variables

x^2+0.1x-0.9975=0

a = 1; b = 0.1; c = -0.9975;
Δ = b2-4ac
Δ = 0.12-4·1·(-0.9975)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.1)-2}{2*1}=\frac{-2.1}{2}
=-1+0.1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.1)+2}{2*1}=\frac{1.9}{2}
=1/1.05263157895
$