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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(12(x+12)), so:

12(x+12)

We multiply parentheses

12x+144

Back to the equation:

-(12x+144)


We get rid of parentheses

x^2-12x-144+16=0

We add all the numbers together, and all the variables

x^2-12x-128=0

a = 1; b = -12; c = -128;
Δ = b2-4ac
Δ = -122-4·1·(-128)
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{41}}{2*1}=\frac{12-4\sqrt{41}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{41}}{2*1}=\frac{12+4\sqrt{41}}{2}
$