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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((x-1)(3x+1)), so:

(x-1)(3x+1)

We multiply parentheses ..

(+3x^2+x-3x-1)

We get rid of parentheses

3x^2+x-3x-1

We add all the numbers together, and all the variables

3x^2-2x-1

Back to the equation:

-(3x^2-2x-1)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-2x^2+8x+10=0

a = -2; b = 8; c = +10;
Δ = b2-4ac
Δ = 82-4·(-2)·10
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{-(8)-12}{2*-2}=\frac{-20}{-4}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12}{2*-2}=\frac{4}{-4}
=-1
$