6(x+2)=-18x+6(x2-3)

Solution for 6(x+2)=-18x+6(x2-3) equation:

6(x+2)=-18x+6(x2-3)

We move all terms to the left:

6(x+2)-(-18x+6(x2-3))=0

We add all the numbers together, and all the variables

-(-18x+6(+x^2-3))+6(x+2)=0

We multiply parentheses

-(-18x+6(+x^2-3))+6x+12=0


We calculate terms in parentheses: -(-18x+6(+x^2-3)), so:

-18x+6(+x^2-3)

determiningTheFunctionDomain
6(+x^2-3)-18x

We multiply parentheses

6x^2-18x-18

Back to the equation:

-(6x^2-18x-18)


We add all the numbers together, and all the variables

6x-(6x^2-18x-18)+12=0

We get rid of parentheses

-6x^2+6x+18x+18+12=0

We add all the numbers together, and all the variables

-6x^2+24x+30=0

a = -6; b = 24; c = +30;
Δ = b2-4ac
Δ = 242-4·(-6)·30
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-36}{2*-6}=\frac{-60}{-12}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+36}{2*-6}=\frac{12}{-12}
=-1
$