x=x^2+0.5x+0.04

Solution for x=x^2+0.5x+0.04 equation:

x=x^2+0.5x+0.04

We move all terms to the left:

x-(x^2+0.5x+0.04)=0

We get rid of parentheses

-x^2+x-0.5x-0.04=0

We add all the numbers together, and all the variables

-1x^2+0.5x-0.04=0

a = -1; b = 0.5; c = -0.04;
Δ = b2-4ac
Δ = 0.52-4·(-1)·(-0.04)
Δ = 0.09
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{-(0.5)-\sqrt{0.09}}{2*-1}=\frac{-0.5-\sqrt{0.09}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.09}}{2*-1}=\frac{-0.5+\sqrt{0.09}}{-2}
$