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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

6x+3x(+/3)-9=0

We multiply parentheses

3x^2+6x-9=0

a = 3; b = 6; c = -9;
Δ = b2-4ac
Δ = 62-4·3·(-9)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12}{2*3}=\frac{-18}{6}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+12}{2*3}=\frac{6}{6}
=1
$