0.1x^2+0.3x=0.1

Solution for 0.1x^2+0.3x=0.1 equation:

0.1x^2+0.3x=0.1

We move all terms to the left:

0.1x^2+0.3x-(0.1)=0

We add all the numbers together, and all the variables

0.1x^2+0.3x-0.1=0

a = 0.1; b = 0.3; c = -0.1;
Δ = b2-4ac
Δ = 0.32-4·0.1·(-0.1)
Δ = 0.13
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{-(0.3)-\sqrt{0.13}}{2*0.1}=\frac{-0.3-\sqrt{0.13}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.3)+\sqrt{0.13}}{2*0.1}=\frac{-0.3+\sqrt{0.13}}{0.2}
$