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solve the following. Use interval notation when appropriate: x^2-7x=12
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and \[\sqrt{3x-2}=7\]
\[\sqrt{3x-2}=7\] Square both sides to get rid of the square root on that one side \[3x-2=7^2\]
3x-2=49 now just add 2 on both sides and then after that divide by 3 on both sides
how about \[x ^{2}-7x+12=0\]
\[12=3(4)=-3(-4) ; -7=-3+(-4)\] so this means you can factor it
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how do I go from the equation I had to what you have?
\[(x-4)(x-3)=0\]
thank you!
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