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

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

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

We move all terms to the left:

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


Domain of the equation: 3)(2x-5)!=0

x∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

3x-(+2/3)(2x-5)+7=0

We multiply parentheses ..

-(+4x^2+2/3*-5)+3x+7=0

We multiply all the terms by the denominator


-(+4x^2+2+3x*3*-5)+7*3*-5)=0

We add all the numbers together, and all the variables

-(+4x^2+2+3x*3*-5)=0

We get rid of parentheses

-4x^2-3x*3*-2+5=0

We add all the numbers together, and all the variables

-4x^2-3x*3*+3=0

Wy multiply elements

-4x^2-9x^2+3=0

We add all the numbers together, and all the variables

-13x^2+3=0

a = -13; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-13)·3
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{39}}{2*-13}=\frac{0-2\sqrt{39}}{-26}
=-\frac{2\sqrt{39}}{-26}
=-\frac{\sqrt{39}}{-13}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{39}}{2*-13}=\frac{0+2\sqrt{39}}{-26}
=\frac{2\sqrt{39}}{-26}
=\frac{\sqrt{39}}{-13}
$