8x(4x-3)=4(6x+4)

Solution for 8x(4x-3)=4(6x+4) equation:

8x(4x-3)=4(6x+4)

We move all terms to the left:

8x(4x-3)-(4(6x+4))=0

We multiply parentheses

32x^2-24x-(4(6x+4))=0


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

4(6x+4)

We multiply parentheses

24x+16

Back to the equation:

-(24x+16)


We get rid of parentheses

32x^2-24x-24x-16=0

We add all the numbers together, and all the variables

32x^2-48x-16=0

a = 32; b = -48; c = -16;
Δ = b2-4ac
Δ = -482-4·32·(-16)
Δ = 4352
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:

$\sqrt{\Delta}=\sqrt{4352}=\sqrt{256*17}=\sqrt{256}*\sqrt{17}=16\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16\sqrt{17}}{2*32}=\frac{48-16\sqrt{17}}{64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16\sqrt{17}}{2*32}=\frac{48+16\sqrt{17}}{64}
$