(1/6)(2x+1)=x+3

Solution for (1/6)(2x+1)=x+3 equation:

(1/6)(2x+1)=x+3

We move all terms to the left:

(1/6)(2x+1)-(x+3)=0


Domain of the equation: 6)(2x+1)!=0

x∈R


We add all the numbers together, and all the variables

(+1/6)(2x+1)-(x+3)=0

We get rid of parentheses

(+1/6)(2x+1)-x-3=0

We multiply parentheses ..

(+2x^2+1/6*1)-x-3=0

We multiply all the terms by the denominator


(+2x^2+1-x*6*1)-3*6*1)=0

We add all the numbers together, and all the variables

(+2x^2+1-x*6*1)=0

We get rid of parentheses

2x^2-x*6*1+1=0

Wy multiply elements

2x^2-6x*1+1=0

Wy multiply elements

2x^2-6x+1=0

a = 2; b = -6; c = +1;
Δ = b2-4ac
Δ = -62-4·2·1
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{7}}{2*2}=\frac{6-2\sqrt{7}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{7}}{2*2}=\frac{6+2\sqrt{7}}{4}
$