3x(2x+4)=2x(20+x)

Solution for 3x(2x+4)=2x(20+x) equation:

3x(2x+4)=2x(20+x)

We move all terms to the left:

3x(2x+4)-(2x(20+x))=0

We add all the numbers together, and all the variables

3x(2x+4)-(2x(x+20))=0

We multiply parentheses

6x^2+12x-(2x(x+20))=0


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

2x(x+20)

We multiply parentheses

2x^2+40x

Back to the equation:

-(2x^2+40x)


We get rid of parentheses

6x^2-2x^2+12x-40x=0

We add all the numbers together, and all the variables

4x^2-28x=0

a = 4; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·4·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*4}=\frac{0}{8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*4}=\frac{56}{8}
=7
$