You can turn this equation into an equation that is quadratic in form by multiplying both sides by ex. That is,
ex + 45e-x - 14 = 0
ex(ex + 45e-x - 14) = ex⋅0
Distribute and add exponents in the above equation to get
e2x + 45e0 - 14ex = 0
e2x + 45 - 14ex = 0 (using the fact that e0 = 1.)
e2x - 14ex + 45 = 0.
Now substitute u = ex. Then, since u2 = (ex)2 = e2x, the above equation can be rewritten as
u2 - 14u + 45 = 0.
(Since the original equation can be turned into a quadratic equation by using an appropriate substitution, we say that the original equation is quadratic in form.) Now it is easy to solve the last equation by factoring.
(u - 9)(u - 5) = 0
u - 9 = 0 or u - 5 = 0
u = 9 or u = 5.
To complete the solution, we substitute ex back in for u, and then take the natural log of each side.
ex = 9 or ex = 5.
Ln(ex) = Ln(9) or Ln(ex) = Ln(5)
x = Ln(9) or x = Ln(5)
x ≅ 2.1972 or x ≅ 1.6094 (rounded to 4 decimal places.)
Hope that helps! Let me know if you need any further help.