3+(x-7)(3x-2)=(x+2)(x-2)-(x+3)(x-3)-3x

Solution for 3+(x-7)(3x-2)=(x+2)(x-2)-(x+3)(x-3)-3x equation:

3+(x-7)(3x-2)=(x+2)(x-2)-(x+3)(x-3)-3x

We move all terms to the left:

3+(x-7)(3x-2)-((x+2)(x-2)-(x+3)(x-3)-3x)=0

We use the square of the difference formula

x^2+(x-7)(3x-2)+4+3=0

We multiply parentheses ..

x^2+(+3x^2-2x-21x+14)+4+3=0

We add all the numbers together, and all the variables

x^2+(+3x^2-2x-21x+14)+7=0

We get rid of parentheses

x^2+3x^2-2x-21x+14+7=0

We add all the numbers together, and all the variables

4x^2-23x+21=0

a = 4; b = -23; c = +21;
Δ = b2-4ac
Δ = -232-4·4·21
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-\sqrt{193}}{2*4}=\frac{23-\sqrt{193}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{193}}{2*4}=\frac{23+\sqrt{193}}{8}
$