(2/3)(9x-6)=4x+10

Solution for (2/3)(9x-6)=4x+10 equation:

(2/3)(9x-6)=4x+10

We move all terms to the left:

(2/3)(9x-6)-(4x+10)=0


Domain of the equation: 3)(9x-6)!=0

x∈R


We add all the numbers together, and all the variables

(+2/3)(9x-6)-(4x+10)=0

We get rid of parentheses

(+2/3)(9x-6)-4x-10=0

We multiply parentheses ..

(+18x^2+2/3*-6)-4x-10=0

We multiply all the terms by the denominator


(+18x^2+2-4x*3*-6)-10*3*-6)=0

We add all the numbers together, and all the variables

(+18x^2+2-4x*3*-6)=0

We get rid of parentheses

18x^2-4x*3*+2-6=0

We add all the numbers together, and all the variables

18x^2-4x*3*-4=0

Wy multiply elements

18x^2-12x^2-4=0

We add all the numbers together, and all the variables

6x^2-4=0

a = 6; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·6·(-4)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*6}=\frac{0-4\sqrt{6}}{12}
=-\frac{4\sqrt{6}}{12}
=-\frac{\sqrt{6}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*6}=\frac{0+4\sqrt{6}}{12}
=\frac{4\sqrt{6}}{12}
=\frac{\sqrt{6}}{3}
$