0.1x^2+0.2x+0.3=0.8x

Solution for 0.1x^2+0.2x+0.3=0.8x equation:

0.1x^2+0.2x+0.3=0.8x

We move all terms to the left:

0.1x^2+0.2x+0.3-(0.8x)=0

We add all the numbers together, and all the variables

0.1x^2+0.2x-(+0.8x)+0.3=0

We get rid of parentheses

0.1x^2+0.2x-0.8x+0.3=0

We add all the numbers together, and all the variables

0.1x^2-0.6x+0.3=0

a = 0.1; b = -0.6; c = +0.3;
Δ = b2-4ac
Δ = -0.62-4·0.1·0.3
Δ = 0.24
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{-(-0.6)-\sqrt{0.24}}{2*0.1}=\frac{0.6-\sqrt{0.24}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.6)+\sqrt{0.24}}{2*0.1}=\frac{0.6+\sqrt{0.24}}{0.2}
$