(3x-1)(x+3)=(x+3)(x+3)

Solution for (3x-1)(x+3)=(x+3)(x+3) equation:

(3x-1)(x+3)=(x+3)(x+3)

We move all terms to the left:

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

We multiply parentheses ..

(+3x^2+9x-1x-3)-((x+3)(x+3))=0


We calculate terms in parentheses: -((x+3)(x+3)), so:

(x+3)(x+3)

We multiply parentheses ..

(+x^2+3x+3x+9)

We get rid of parentheses

x^2+3x+3x+9

We add all the numbers together, and all the variables

x^2+6x+9

Back to the equation:

-(x^2+6x+9)


We get rid of parentheses

3x^2-x^2+9x-1x-6x-3-9=0

We add all the numbers together, and all the variables

2x^2+2x-12=0

a = 2; b = 2; c = -12;
Δ = b2-4ac
Δ = 22-4·2·(-12)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*2}=\frac{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*2}=\frac{8}{4}
=2
$