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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+2x^2-16x-4x+32))+2442=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^2-20x+32

Back to the equation:

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


We get rid of parentheses

-2x^2+20x-32+2442=0

We add all the numbers together, and all the variables

-2x^2+20x+2410=0

a = -2; b = 20; c = +2410;
Δ = b2-4ac
Δ = 202-4·(-2)·2410
Δ = 19680
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:

$\sqrt{\Delta}=\sqrt{19680}=\sqrt{16*1230}=\sqrt{16}*\sqrt{1230}=4\sqrt{1230}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{1230}}{2*-2}=\frac{-20-4\sqrt{1230}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{1230}}{2*-2}=\frac{-20+4\sqrt{1230}}{-4}
$