(2x-5)-(2/3x+3)=12

Solution for (2x-5)-(2/3x+3)=12 equation:

(2x-5)-(2/3x+3)=12

We move all terms to the left:

(2x-5)-(2/3x+3)-(12)=0


Domain of the equation: 3x+3)!=0

x∈R


We get rid of parentheses

2x-2/3x-5-3-12=0

We multiply all the terms by the denominator


2x*3x-5*3x-3*3x-12*3x-2=0

Wy multiply elements

6x^2-15x-9x-36x-2=0

We add all the numbers together, and all the variables

6x^2-60x-2=0

a = 6; b = -60; c = -2;
Δ = b2-4ac
Δ = -602-4·6·(-2)
Δ = 3648
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:

$\sqrt{\Delta}=\sqrt{3648}=\sqrt{64*57}=\sqrt{64}*\sqrt{57}=8\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-8\sqrt{57}}{2*6}=\frac{60-8\sqrt{57}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+8\sqrt{57}}{2*6}=\frac{60+8\sqrt{57}}{12}
$