Please expand the brackets in the following equation to get a quadratic equation. Then, please show using the quadratic formula that the solutions to the equation are x=3 and x=5. Here is the starting equation: (x-3)(x-5)=0
(X-3)(x-5)=0Use F.O.I.LFirst x multiplied by x gives x2outer x multiplied by -5 gives -5xinner x multiplied by -3 gives -3xlast -5x-3 gives +15combining we get x2-8x+15=0the quadratic equation ax2+bx+c=0 has solutions (-b+-sqrt(b2-4ac))/2a, or (8+-sqrt(64-60))/2=(8+-2)/2=5 and 3.
JG
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