Question

How do I solve 2x^x-5=0 using the quadradic formula?

Answer 1

In order to use the quadratic formula, you have to make sure that
your quadratic equation is in standard form, and that you know exactly
what the 'A', 'B' and 'C' numbers are before you take and plug them
into the formula.

You said your equation is            2 x^x  -  5  =  0 .

At this point, I must ask you a very serious question: 

                                       What do you mean by ' x^x ' ?

If you mean  '  x  times  x  ", then we can write that as ' x² ' and continue.

If you mean  ' x raised to the x power ', then this thing is not a
quadratic equation.  It's a beast of a different kind, much more
complicated than a quadratic equation.  It can't be solved with
the quadratic formula, and we can stop right here.

So I have to assume that your equation is

                                               2x² -  5  =  0

You need it in the form of    Ax² + Bx + C  =  0 .

That's easy.  Just write      (2)x² + (0)x + (-5)  =  0

                       Then          A=2      B=0      C= -5 .

And now you can stuff A, B, and C directly into the quadratic formula,
and bada-bing, out come the two solutions:

                 x = +1.5811...
                 x = - 1.5811... 

Answer 2

So in comments below
2x^2+x-5=0
if you have
ax^2+bx+c=0 then
x=$\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}$

a=2
b=1
c=-5

subsitue
x=$\frac{-1+/- \sqrt{1^{2}-4(2)(-5)} }{2(2)}$
x=$\frac{-1+/- \sqrt{1+40} }{4}$

so x=$\frac{-1+ \sqrt{1+40} }{4}$ or $\frac{-1- \sqrt{1+40} }{4}$

aprox
x=-1.85078 or 1.35078