(x-5)=2x(10x-5x+20)

Solution for (x-5)=2x(10x-5x+20) equation:

(x-5)=2x(10x-5x+20)

We move all terms to the left:

(x-5)-(2x(10x-5x+20))=0

We add all the numbers together, and all the variables

(x-5)-(2x(5x+20))=0

We get rid of parentheses

x-(2x(5x+20))-5=0


We calculate terms in parentheses: -(2x(5x+20)), so:

2x(5x+20)

We multiply parentheses

10x^2+40x

Back to the equation:

-(10x^2+40x)


We get rid of parentheses

-10x^2+x-40x-5=0

We add all the numbers together, and all the variables

-10x^2-39x-5=0

a = -10; b = -39; c = -5;
Δ = b2-4ac
Δ = -392-4·(-10)·(-5)
Δ = 1321
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{-(-39)-\sqrt{1321}}{2*-10}=\frac{39-\sqrt{1321}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{1321}}{2*-10}=\frac{39+\sqrt{1321}}{-20}
$