x-(2/5x)+21=403

Solution for x-(2/5x)+21=403 equation:

x-(2/5x)+21=403

We move all terms to the left:

x-(2/5x)+21-(403)=0


Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

x-(+2/5x)+21-403=0

We add all the numbers together, and all the variables

x-(+2/5x)-382=0

We get rid of parentheses

x-2/5x-382=0

We multiply all the terms by the denominator


x*5x-382*5x-2=0

Wy multiply elements

5x^2-1910x-2=0

a = 5; b = -1910; c = -2;
Δ = b2-4ac
Δ = -19102-4·5·(-2)
Δ = 3648140
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3648140}=\sqrt{4*912035}=\sqrt{4}*\sqrt{912035}=2\sqrt{912035}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1910)-2\sqrt{912035}}{2*5}=\frac{1910-2\sqrt{912035}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1910)+2\sqrt{912035}}{2*5}=\frac{1910+2\sqrt{912035}}{10}
$