3(x+2)(x-2)=(x-4)+8x

Solution for 3(x+2)(x-2)=(x-4)+8x equation:

3(x+2)(x-2)=(x-4)+8x

We move all terms to the left:

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

We use the square of the difference formula

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


We calculate terms in parentheses: -((x-4)+8x), so:

(x-4)+8x

We add all the numbers together, and all the variables

8x+(x-4)

We get rid of parentheses

8x+x-4

We add all the numbers together, and all the variables

9x-4

Back to the equation:

-(9x-4)


We get rid of parentheses

x^2-9x+4-4=0

We add all the numbers together, and all the variables

x^2-9x=0

a = 1; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·1·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*1}=\frac{18}{2}
=9
$