4x(x+4)=(3x+8)(x+4)

Solution for 4x(x+4)=(3x+8)(x+4) equation:

4x(x+4)=(3x+8)(x+4)

We move all terms to the left:

4x(x+4)-((3x+8)(x+4))=0

We multiply parentheses

4x^2+16x-((3x+8)(x+4))=0

We multiply parentheses ..

4x^2-((+3x^2+12x+8x+32))+16x=0


We calculate terms in parentheses: -((+3x^2+12x+8x+32)), so:

(+3x^2+12x+8x+32)

We get rid of parentheses

3x^2+12x+8x+32

We add all the numbers together, and all the variables

3x^2+20x+32

Back to the equation:

-(3x^2+20x+32)


We add all the numbers together, and all the variables

4x^2+16x-(3x^2+20x+32)=0

We get rid of parentheses

4x^2-3x^2+16x-20x-32=0

We add all the numbers together, and all the variables

x^2-4x-32=0

a = 1; b = -4; c = -32;
Δ = b2-4ac
Δ = -42-4·1·(-32)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*1}=\frac{16}{2}
=8
$