x^2+0.2x+0.01=0.23(0.5-x)

Solution for x^2+0.2x+0.01=0.23(0.5-x) equation:

x^2+0.2x+0.01=0.23(0.5-x)

We move all terms to the left:

x^2+0.2x+0.01-(0.23(0.5-x))=0

We add all the numbers together, and all the variables

x^2+0.2x-(0.23(-1x+0.5))+0.01=0


We calculate terms in parentheses: -(0.23(-1x+0.5)), so:

0.23(-1x+0.5)

We multiply parentheses

-0.23x+0.115

Back to the equation:

-(-0.23x+0.115)


We get rid of parentheses

x^2+0.2x+0.23x-0.115+0.01=0

We add all the numbers together, and all the variables

x^2+0.43x-0.105=0

a = 1; b = 0.43; c = -0.105;
Δ = b2-4ac
Δ = 0.432-4·1·(-0.105)
Δ = 0.6049
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{-(0.43)-\sqrt{0.6049}}{2*1}=\frac{-0.43-\sqrt{0.6049}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.43)+\sqrt{0.6049}}{2*1}=\frac{-0.43+\sqrt{0.6049}}{2}
$