3x(6x-4)+18=10+3(2+4x)

Solution for 3x(6x-4)+18=10+3(2+4x) equation:

3x(6x-4)+18=10+3(2+4x)

We move all terms to the left:

3x(6x-4)+18-(10+3(2+4x))=0

We add all the numbers together, and all the variables

3x(6x-4)-(10+3(4x+2))+18=0

We multiply parentheses

18x^2-12x-(10+3(4x+2))+18=0


We calculate terms in parentheses: -(10+3(4x+2)), so:

10+3(4x+2)

determiningTheFunctionDomain
3(4x+2)+10

We multiply parentheses

12x+6+10

We add all the numbers together, and all the variables

12x+16

Back to the equation:

-(12x+16)


We get rid of parentheses

18x^2-12x-12x-16+18=0

We add all the numbers together, and all the variables

18x^2-24x+2=0

a = 18; b = -24; c = +2;
Δ = b2-4ac
Δ = -242-4·18·2
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-12\sqrt{3}}{2*18}=\frac{24-12\sqrt{3}}{36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+12\sqrt{3}}{2*18}=\frac{24+12\sqrt{3}}{36}
$