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The browser used to post questions on this site is utter rubbish: it will not accept most mathematical symbols and so it is difficult to be sure what the question is. What we can see of the equation is "e(2x)ex plus 2". This is an expression - not an equation - and expressions do not have solutions! However, supposing the equation is e^2x +/- e^x + 2 = 0Then, if you substitute y = e^x you get a quadratic equation in y. Whether the second term of the quadratic is positive or negative, the discriminant is 1 - 8 = -7. So there are no real roots - not one.

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