2^4=((x^2)+1)/(x-3)

Solution for 2^4=((x^2)+1)/(x-3) equation:

2^4=((x^2)+1)/(x-3)

We move all terms to the left:

2^4-(((x^2)+1)/(x-3))=0


Domain of the equation: (x-3))!=0

x∈R


We add all the numbers together, and all the variables

-((x^2+1)/(x-3))+16=0

We multiply all the terms by the denominator


-((x^2+1)+16*(x-3))=0


We calculate terms in parentheses: -((x^2+1)+16*(x-3)), so:

(x^2+1)+16*(x-3)

We multiply parentheses

(x^2+1)+16x-48

We get rid of parentheses

x^2+16x+1-48

We add all the numbers together, and all the variables

x^2+16x-47

Back to the equation:

-(x^2+16x-47)


We get rid of parentheses

-x^2-16x+47=0

We add all the numbers together, and all the variables

-1x^2-16x+47=0

a = -1; b = -16; c = +47;
Δ = b2-4ac
Δ = -162-4·(-1)·47
Δ = 444
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:

$\sqrt{\Delta}=\sqrt{444}=\sqrt{4*111}=\sqrt{4}*\sqrt{111}=2\sqrt{111}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{111}}{2*-1}=\frac{16-2\sqrt{111}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{111}}{2*-1}=\frac{16+2\sqrt{111}}{-2}
$