3(2x2)-4(x+6)=3x+8+x

Solution for 3(2x2)-4(x+6)=3x+8+x equation:

3(2x^2)-4(x+6)=3x+8+x

We move all terms to the left:

3(2x^2)-4(x+6)-(3x+8+x)=0

We add all the numbers together, and all the variables

32x^2-4(x+6)-(4x+8)=0

We multiply parentheses

32x^2-4x-(4x+8)-24=0

We get rid of parentheses

32x^2-4x-4x-8-24=0

We add all the numbers together, and all the variables

32x^2-8x-32=0

a = 32; b = -8; c = -32;
Δ = b2-4ac
Δ = -82-4·32·(-32)
Δ = 4160
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:

$\sqrt{\Delta}=\sqrt{4160}=\sqrt{64*65}=\sqrt{64}*\sqrt{65}=8\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{65}}{2*32}=\frac{8-8\sqrt{65}}{64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{65}}{2*32}=\frac{8+8\sqrt{65}}{64}
$