-2/3x+12=-10+x

Solution for -2/3x+12=-10+x equation:

-2/3x+12=-10+x

We move all terms to the left:

-2/3x+12-(-10+x)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

-2/3x-(x-10)+12=0

We get rid of parentheses

-2/3x-x+10+12=0

We multiply all the terms by the denominator


-x*3x+10*3x+12*3x-2=0

Wy multiply elements

-3x^2+30x+36x-2=0

We add all the numbers together, and all the variables

-3x^2+66x-2=0

a = -3; b = 66; c = -2;
Δ = b2-4ac
Δ = 662-4·(-3)·(-2)
Δ = 4332
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:

$\sqrt{\Delta}=\sqrt{4332}=\sqrt{1444*3}=\sqrt{1444}*\sqrt{3}=38\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(66)-38\sqrt{3}}{2*-3}=\frac{-66-38\sqrt{3}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(66)+38\sqrt{3}}{2*-3}=\frac{-66+38\sqrt{3}}{-6}
$