24x+12=2(12x+6)5x-20=3(x-4)

Solution for 24x+12=2(12x+6)5x-20=3(x-4) equation:

24x+12=2(12x+6)5x-20=3(x-4)

We move all terms to the left:

24x+12-(2(12x+6)5x-20)=0


We calculate terms in parentheses: -(2(12x+6)5x-20), so:

2(12x+6)5x-20

We multiply parentheses

120x^2+60x-20

Back to the equation:

-(120x^2+60x-20)


We get rid of parentheses

-120x^2+24x-60x+20+12=0

We add all the numbers together, and all the variables

-120x^2-36x+32=0

a = -120; b = -36; c = +32;
Δ = b2-4ac
Δ = -362-4·(-120)·32
Δ = 16656
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:

$\sqrt{\Delta}=\sqrt{16656}=\sqrt{16*1041}=\sqrt{16}*\sqrt{1041}=4\sqrt{1041}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{1041}}{2*-120}=\frac{36-4\sqrt{1041}}{-240}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{1041}}{2*-120}=\frac{36+4\sqrt{1041}}{-240}
$