(x^2-16)=(3x+12)(x+3)

Solution for (x^2-16)=(3x+12)(x+3) equation:

(x^2-16)=(3x+12)(x+3)

We move all terms to the left:

(x^2-16)-((3x+12)(x+3))=0

We get rid of parentheses

x^2-((3x+12)(x+3))-16=0

We multiply parentheses ..

x^2-((+3x^2+9x+12x+36))-16=0


We calculate terms in parentheses: -((+3x^2+9x+12x+36)), so:

(+3x^2+9x+12x+36)

We get rid of parentheses

3x^2+9x+12x+36

We add all the numbers together, and all the variables

3x^2+21x+36

Back to the equation:

-(3x^2+21x+36)


We get rid of parentheses

x^2-3x^2-21x-36-16=0

We add all the numbers together, and all the variables

-2x^2-21x-52=0

a = -2; b = -21; c = -52;
Δ = b2-4ac
Δ = -212-4·(-2)·(-52)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-5}{2*-2}=\frac{16}{-4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+5}{2*-2}=\frac{26}{-4}
=-6+1/2
$