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Answer to Question #3661 in Algebra for tami

Question #3661
Solve 2e2x + 19ex-27=0.
Moderate step by step would be great! Thanks!
Expert's answer
Denote ex as t, thus the equation is
2t2 +19t -27 = 0
D = 192 + 4*2*27 = 577
t1 = (-19 + √577)/4
t2 = (-19 - √577)/4 < 0 - doesnt's satisfy the conditions, as ex >=0 at any x.

ex = (-19 + √577)/4, x = ln(-19 + √577) - ln4

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