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Solve by completing the square x^2+2x-3=0
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We can use (x-1)(x+3) = 0 Therefore, x = 1, x = -3
Or we can also do like this: x^2 + 2x - 3=0 x^2 +2x + 1 -4=0(I just wrote 1-4 instead of -3) (x+1)^2-4=0 (x+1)^2=4 \[(x+1)=\pm2\] x=1 or x=-3
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