6+12^2+2x(3-6x)+8=-2(3+2x)

Solution for 6+12^2+2x(3-6x)+8=-2(3+2x) equation:

6+12^2+2x(3-6x)+8=-2(3+2x)

We move all terms to the left:

6+12^2+2x(3-6x)+8-(-2(3+2x))=0

We add all the numbers together, and all the variables

2x(-6x+3)-(-2(2x+3))+6+8+12^2=0

We add all the numbers together, and all the variables

2x(-6x+3)-(-2(2x+3))+158=0

We multiply parentheses

-12x^2+6x-(-2(2x+3))+158=0


We calculate terms in parentheses: -(-2(2x+3)), so:

-2(2x+3)

We multiply parentheses

-4x-6

Back to the equation:

-(-4x-6)


We get rid of parentheses

-12x^2+6x+4x+6+158=0

We add all the numbers together, and all the variables

-12x^2+10x+164=0

a = -12; b = 10; c = +164;
Δ = b2-4ac
Δ = 102-4·(-12)·164
Δ = 7972
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:

$\sqrt{\Delta}=\sqrt{7972}=\sqrt{4*1993}=\sqrt{4}*\sqrt{1993}=2\sqrt{1993}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{1993}}{2*-12}=\frac{-10-2\sqrt{1993}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{1993}}{2*-12}=\frac{-10+2\sqrt{1993}}{-24}
$