2x^-12x+32=x^2

Solution for 2x^-12x+32=x^2 equation:

2x^-12x+32=x^2

We move all terms to the left:

2x^-12x+32-(x^2)=0

determiningTheFunctionDomain
-x^2+2x^-12x+32=0

We add all the numbers together, and all the variables

-1x^2-10x+32=0

a = -1; b = -10; c = +32;
Δ = b2-4ac
Δ = -102-4·(-1)·32
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{57}}{2*-1}=\frac{10-2\sqrt{57}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{57}}{2*-1}=\frac{10+2\sqrt{57}}{-2}
$