5x+(2x-(3x-20))=x(400-2x)

Solution for 5x+(2x-(3x-20))=x(400-2x) equation:

5x+(2x-(3x-20))=x(400-2x)

We move all terms to the left:

5x+(2x-(3x-20))-(x(400-2x))=0

We add all the numbers together, and all the variables

5x+(2x-(3x-20))-(x(-2x+400))=0


We calculate terms in parentheses: +(2x-(3x-20)), so:

2x-(3x-20)

We get rid of parentheses

2x-3x+20

We add all the numbers together, and all the variables

-1x+20

Back to the equation:

+(-1x+20)



We calculate terms in parentheses: -(x(-2x+400)), so:

x(-2x+400)

We multiply parentheses

-2x^2+400x

Back to the equation:

-(-2x^2+400x)


We get rid of parentheses

2x^2-400x+5x-1x+20=0

We add all the numbers together, and all the variables

2x^2-396x+20=0

a = 2; b = -396; c = +20;
Δ = b2-4ac
Δ = -3962-4·2·20
Δ = 156656
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:

$\sqrt{\Delta}=\sqrt{156656}=\sqrt{16*9791}=\sqrt{16}*\sqrt{9791}=4\sqrt{9791}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-396)-4\sqrt{9791}}{2*2}=\frac{396-4\sqrt{9791}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-396)+4\sqrt{9791}}{2*2}=\frac{396+4\sqrt{9791}}{4}
$