(42)(24)=(2x-4)(x+8)

Solution for (42)(24)=(2x-4)(x+8) equation:

(42)(24)=(2x-4)(x+8)

We move all terms to the left:

(42)(24)-((2x-4)(x+8))=0

We multiply parentheses ..

-((+2x^2+16x-4x-32))+4224=0


We calculate terms in parentheses: -((+2x^2+16x-4x-32)), so:

(+2x^2+16x-4x-32)

We get rid of parentheses

2x^2+16x-4x-32

We add all the numbers together, and all the variables

2x^2+12x-32

Back to the equation:

-(2x^2+12x-32)


We get rid of parentheses

-2x^2-12x+32+4224=0

We add all the numbers together, and all the variables

-2x^2-12x+4256=0

a = -2; b = -12; c = +4256;
Δ = b2-4ac
Δ = -122-4·(-2)·4256
Δ = 34192
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:

$\sqrt{\Delta}=\sqrt{34192}=\sqrt{16*2137}=\sqrt{16}*\sqrt{2137}=4\sqrt{2137}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{2137}}{2*-2}=\frac{12-4\sqrt{2137}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{2137}}{2*-2}=\frac{12+4\sqrt{2137}}{-4}
$