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

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

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

We move all terms to the left:

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


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

x∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We multiply all the terms by the denominator


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

We get rid of parentheses

3x^2-3x*2*3+1=0

Wy multiply elements

3x^2-18x*3+1=0

Wy multiply elements

3x^2-54x+1=0

a = 3; b = -54; c = +1;
Δ = b2-4ac
Δ = -542-4·3·1
Δ = 2904
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:

$\sqrt{\Delta}=\sqrt{2904}=\sqrt{484*6}=\sqrt{484}*\sqrt{6}=22\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-54)-22\sqrt{6}}{2*3}=\frac{54-22\sqrt{6}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-54)+22\sqrt{6}}{2*3}=\frac{54+22\sqrt{6}}{6}
$