x^2-(2/3)x+(1/9)=0

Solution for x^2-(2/3)x+(1/9)=0 equation:

x^2-(2/3)x+(1/9)=0


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

x^2-(+2/3)x+(+1/9)=0

We multiply parentheses

x^2-2x^2+(+1/9)=0

We get rid of parentheses

x^2-2x^2+1/9=0

We multiply all the terms by the denominator


x^2*9-2x^2*9+1=0

Wy multiply elements

9x^2-18x^2+1=0

We add all the numbers together, and all the variables

-9x^2+1=0

a = -9; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-9)·1
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*-9}=\frac{-6}{-18}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*-9}=\frac{6}{-18}
=-1/3
$