(x-12)(x+12)=(3-x)(x+8)

Solution for (x-12)(x+12)=(3-x)(x+8) equation:

(x-12)(x+12)=(3-x)(x+8)

We move all terms to the left:

(x-12)(x+12)-((3-x)(x+8))=0

We add all the numbers together, and all the variables

(x-12)(x+12)-((-1x+3)(x+8))=0

We use the square of the difference formula

x^2-((-1x+3)(x+8))-144=0

We multiply parentheses ..

x^2-((-1x^2-8x+3x+24))-144=0


We calculate terms in parentheses: -((-1x^2-8x+3x+24)), so:

(-1x^2-8x+3x+24)

We get rid of parentheses

-1x^2-8x+3x+24

We add all the numbers together, and all the variables

-1x^2-5x+24

Back to the equation:

-(-1x^2-5x+24)


We get rid of parentheses

x^2+1x^2+5x-24-144=0

We add all the numbers together, and all the variables

2x^2+5x-168=0

a = 2; b = 5; c = -168;
Δ = b2-4ac
Δ = 52-4·2·(-168)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-37}{2*2}=\frac{-42}{4}
=-10+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+37}{2*2}=\frac{32}{4}
=8
$