8(x+7)=6(x-4)5x-20=3(x-4)24x+12=2(12x+6)

Solution for 8(x+7)=6(x-4)5x-20=3(x-4)24x+12=2(12x+6) equation:

8(x+7)=6(x-4)5x-20=3(x-4)24x+12=2(12x+6)

We move all terms to the left:

8(x+7)-(6(x-4)5x-20)=0

We multiply parentheses

8x-(6(x-4)5x-20)+56=0


We calculate terms in parentheses: -(6(x-4)5x-20), so:

6(x-4)5x-20

We multiply parentheses

30x^2-120x-20

Back to the equation:

-(30x^2-120x-20)


We get rid of parentheses

-30x^2+8x+120x+20+56=0

We add all the numbers together, and all the variables

-30x^2+128x+76=0

a = -30; b = 128; c = +76;
Δ = b2-4ac
Δ = 1282-4·(-30)·76
Δ = 25504
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:

$\sqrt{\Delta}=\sqrt{25504}=\sqrt{16*1594}=\sqrt{16}*\sqrt{1594}=4\sqrt{1594}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-4\sqrt{1594}}{2*-30}=\frac{-128-4\sqrt{1594}}{-60}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+4\sqrt{1594}}{2*-30}=\frac{-128+4\sqrt{1594}}{-60}
$