what is the solution to x2+x-12=8

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$x^2+x-12=8 \\ \Leftrightarrow x^2+x-20=0 \; \; \; \text {Two numbers which add to give 1 and multiply to give -20 will be} \\ \text{-4 and 5 as they add to make +1 and miltiply giving -20.} \\ = (x-4)(x+5) \\ \text{Therefore the solutions are x=4 or x=-5}$

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