x+(x+20)=3x(x-8)

Solution for x+(x+20)=3x(x-8) equation:

x+(x+20)=3x(x-8)

We move all terms to the left:

x+(x+20)-(3x(x-8))=0

We get rid of parentheses

x+x-(3x(x-8))+20=0


We calculate terms in parentheses: -(3x(x-8)), so:

3x(x-8)

We multiply parentheses

3x^2-24x

Back to the equation:

-(3x^2-24x)


We add all the numbers together, and all the variables

2x-(3x^2-24x)+20=0

We get rid of parentheses

-3x^2+2x+24x+20=0

We add all the numbers together, and all the variables

-3x^2+26x+20=0

a = -3; b = 26; c = +20;
Δ = b2-4ac
Δ = 262-4·(-3)·20
Δ = 916
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:

$\sqrt{\Delta}=\sqrt{916}=\sqrt{4*229}=\sqrt{4}*\sqrt{229}=2\sqrt{229}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{229}}{2*-3}=\frac{-26-2\sqrt{229}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{229}}{2*-3}=\frac{-26+2\sqrt{229}}{-6}
$