(2/3)x-(1/4)x+(1/12)x=3

Solution for (2/3)x-(1/4)x+(1/12)x=3 equation:

(2/3)x-(1/4)x+(1/12)x=3

We move all terms to the left:

(2/3)x-(1/4)x+(1/12)x-(3)=0


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 4)x!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 12)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+2/3)x-(+1/4)x+(+1/12)x-3=0

We multiply parentheses

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

We add all the numbers together, and all the variables

2x^2-3=0

a = 2; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·2·(-3)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*2}=\frac{0-2\sqrt{6}}{4}
=-\frac{2\sqrt{6}}{4}
=-\frac{\sqrt{6}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*2}=\frac{0+2\sqrt{6}}{4}
=\frac{2\sqrt{6}}{4}
=\frac{\sqrt{6}}{2}
$