(39)(24)=(2x-4)(x+6)

Solution for (39)(24)=(2x-4)(x+6) equation:

(39)(24)=(2x-4)(x+6)

We move all terms to the left:

(39)(24)-((2x-4)(x+6))=0

We multiply parentheses ..

-((+2x^2+12x-4x-24))+3924=0


We calculate terms in parentheses: -((+2x^2+12x-4x-24)), so:

(+2x^2+12x-4x-24)

We get rid of parentheses

2x^2+12x-4x-24

We add all the numbers together, and all the variables

2x^2+8x-24

Back to the equation:

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


We get rid of parentheses

-2x^2-8x+24+3924=0

We add all the numbers together, and all the variables

-2x^2-8x+3948=0

a = -2; b = -8; c = +3948;
Δ = b2-4ac
Δ = -82-4·(-2)·3948
Δ = 31648
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:

$\sqrt{\Delta}=\sqrt{31648}=\sqrt{16*1978}=\sqrt{16}*\sqrt{1978}=4\sqrt{1978}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{1978}}{2*-2}=\frac{8-4\sqrt{1978}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{1978}}{2*-2}=\frac{8+4\sqrt{1978}}{-4}
$