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

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

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

We move all terms to the left:

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

We use the square of the difference formula

9x^2-(-(x-3)(x+1))-4=0

We multiply parentheses ..

9x^2-(-(+x^2+x-3x-3))-4=0


We calculate terms in parentheses: -(-(+x^2+x-3x-3)), so:

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

We get rid of parentheses

-x^2-x+3x+3

We add all the numbers together, and all the variables

-1x^2+2x+3

Back to the equation:

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


We get rid of parentheses

9x^2+1x^2-2x-3-4=0

We add all the numbers together, and all the variables

10x^2-2x-7=0

a = 10; b = -2; c = -7;
Δ = b2-4ac
Δ = -22-4·10·(-7)
Δ = 284
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:

$\sqrt{\Delta}=\sqrt{284}=\sqrt{4*71}=\sqrt{4}*\sqrt{71}=2\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{71}}{2*10}=\frac{2-2\sqrt{71}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{71}}{2*10}=\frac{2+2\sqrt{71}}{20}
$