0.2x^2=-0.3x+0.1

Solution for 0.2x^2=-0.3x+0.1 equation:

0.2x^2=-0.3x+0.1

We move all terms to the left:

0.2x^2-(-0.3x+0.1)=0

We get rid of parentheses

0.2x^2+0.3x-0.1=0

a = 0.2; b = 0.3; c = -0.1;
Δ = b2-4ac
Δ = 0.32-4·0.2·(-0.1)
Δ = 0.17
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{0.17}}{2*0.2}=\frac{-0.3-\sqrt{0.17}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.3)+\sqrt{0.17}}{2*0.2}=\frac{-0.3+\sqrt{0.17}}{0.4}
$