x^2+0.2x+0.01=0.5-x

Solution for x^2+0.2x+0.01=0.5-x equation:

x^2+0.2x+0.01=0.5-x

We move all terms to the left:

x^2+0.2x+0.01-(0.5-x)=0

We add all the numbers together, and all the variables

x^2+0.2x-(-1x+0.5)+0.01=0

We get rid of parentheses

x^2+0.2x+1x-0.5+0.01=0

We add all the numbers together, and all the variables

x^2+1.2x-0.49=0

a = 1; b = 1.2; c = -0.49;
Δ = b2-4ac
Δ = 1.22-4·1·(-0.49)
Δ = 3.4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1.2)-\sqrt{3.4}}{2*1}=\frac{-1.2-\sqrt{3.4}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.2)+\sqrt{3.4}}{2*1}=\frac{-1.2+\sqrt{3.4}}{2}
$