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Solve for x and I need to see the steps x^2e^-x-xe^-x-12e^-x=0
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factor out e^-x\[e^{-x}(x^2-x-12)=0\]
\[x^2-x-12=0\]
\[(x-4)(x+3)=0\]
Let e^-x be b bx^2 - bx - 12b = 0 b (x^2 - x - 12) = 0 b ( x + 3) ( x - 4) = 0 x = -3, 4
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