12=(2x+8)(4x-3)

Solution for 12=(2x+8)(4x-3) equation:

12=(2x+8)(4x-3)

We move all terms to the left:

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

We multiply parentheses ..

-((+8x^2-6x+32x-24))+12=0


We calculate terms in parentheses: -((+8x^2-6x+32x-24)), so:

(+8x^2-6x+32x-24)

We get rid of parentheses

8x^2-6x+32x-24

We add all the numbers together, and all the variables

8x^2+26x-24

Back to the equation:

-(8x^2+26x-24)


We get rid of parentheses

-8x^2-26x+24+12=0

We add all the numbers together, and all the variables

-8x^2-26x+36=0

a = -8; b = -26; c = +36;
Δ = b2-4ac
Δ = -262-4·(-8)·36
Δ = 1828
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:

$\sqrt{\Delta}=\sqrt{1828}=\sqrt{4*457}=\sqrt{4}*\sqrt{457}=2\sqrt{457}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{457}}{2*-8}=\frac{26-2\sqrt{457}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{457}}{2*-8}=\frac{26+2\sqrt{457}}{-16}
$