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b²-16b+64=16 SOLVE PL0X
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we have \[b^2-16b+64=16\] subtract both sides by 16 \[b^2-16b+48\] let's find factors of 48 ,which sums up to -48 -12 and -4 -12+(-4)=-16 -12*-4=48 so we have now \[b^2-12b-4b+48\] \[b(b-12)-4(b-12)\] we get finally \[(b-4)(b-12)\] so the roots are b=4 and b=12
woops, i didnt see the 16, lol. im blind xD
thanks
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