x^2-(2/5)x+(1/25)=0

Solution for x^2-(2/5)x+(1/25)=0 equation:

x^2-(2/5)x+(1/25)=0


Domain of the equation: 5)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

x^2-(+2/5)x+(+1/25)=0

We multiply parentheses

x^2-2x^2+(+1/25)=0

We get rid of parentheses

x^2-2x^2+1/25=0

We multiply all the terms by the denominator


x^2*25-2x^2*25+1=0

Wy multiply elements

25x^2-50x^2+1=0

We add all the numbers together, and all the variables

-25x^2+1=0

a = -25; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-25)·1
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10}{2*-25}=\frac{-10}{-50}
=1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*-25}=\frac{10}{-50}
=-1/5
$