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

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

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

We move all terms to the left:

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


Domain of the equation: 2)(2x+4)!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

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

We multiply all the terms by the denominator


(+2x^2+1-x*2*4)-3*2*4)=0

We add all the numbers together, and all the variables

(+2x^2+1-x*2*4)=0

We get rid of parentheses

2x^2-x*2*4+1=0

Wy multiply elements

2x^2-8x*4+1=0

Wy multiply elements

2x^2-32x+1=0

a = 2; b = -32; c = +1;
Δ = b2-4ac
Δ = -322-4·2·1
Δ = 1016
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:

$\sqrt{\Delta}=\sqrt{1016}=\sqrt{4*254}=\sqrt{4}*\sqrt{254}=2\sqrt{254}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-2\sqrt{254}}{2*2}=\frac{32-2\sqrt{254}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+2\sqrt{254}}{2*2}=\frac{32+2\sqrt{254}}{4}
$