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

Solution for 9(x-2)+3x=6(-3+x2) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(6(+x^2-3))+9(x-2)+3x=0

We add all the numbers together, and all the variables

-(6(+x^2-3))+3x+9(x-2)=0

We multiply parentheses

-(6(+x^2-3))+3x+9x-18=0


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

6(+x^2-3)

We multiply parentheses

6x^2-18

Back to the equation:

-(6x^2-18)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-6x^2+12x=0

a = -6; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-6)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-6}=\frac{-24}{-12}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-6}=\frac{0}{-12}
=0
$