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

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

X=(4X-4)(3X+2)

We move all terms to the left:

X-((4X-4)(3X+2))=0

We multiply parentheses ..

-((+12X^2+8X-12X-8))+X=0


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

(+12X^2+8X-12X-8)

We get rid of parentheses

12X^2+8X-12X-8

We add all the numbers together, and all the variables

12X^2-4X-8

Back to the equation:

-(12X^2-4X-8)


We add all the numbers together, and all the variables

X-(12X^2-4X-8)=0

We get rid of parentheses

-12X^2+X+4X+8=0

We add all the numbers together, and all the variables

-12X^2+5X+8=0

a = -12; b = 5; c = +8;
Δ = b2-4ac
Δ = 52-4·(-12)·8
Δ = 409
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{409}}{2*-12}=\frac{-5-\sqrt{409}}{-24}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{409}}{2*-12}=\frac{-5+\sqrt{409}}{-24}
$