(x^2+12)=(6x+16)

Solution for (x^2+12)=(6x+16) equation:

(x^2+12)=(6x+16)

We move all terms to the left:

(x^2+12)-((6x+16))=0

We get rid of parentheses

x^2-((6x+16))+12=0


We calculate terms in parentheses: -((6x+16)), so:

(6x+16)

We get rid of parentheses

6x+16

Back to the equation:

-(6x+16)


We get rid of parentheses

x^2-6x-16+12=0

We add all the numbers together, and all the variables

x^2-6x-4=0

a = 1; b = -6; c = -4;
Δ = b2-4ac
Δ = -62-4·1·(-4)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{13}}{2*1}=\frac{6-2\sqrt{13}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{13}}{2*1}=\frac{6+2\sqrt{13}}{2}
$