(1/3)-(5/3)x+2x^2=0

Solution for (1/3)-(5/3)x+2x^2=0 equation:

(1/3)-(5/3)x+2x^2=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

2x^2-(+5/3)x+(+1/3)=0

We multiply parentheses

2x^2-5x^2+(+1/3)=0

We get rid of parentheses

2x^2-5x^2+1/3=0

We multiply all the terms by the denominator


2x^2*3-5x^2*3+1=0

Wy multiply elements

6x^2-15x^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
$