Question 241663
There are two problems with your question:
1.  We cannot setup an (x,y) table when there is no y in the equation.
2.  You are making reference to a quadratic equation, but there is no quadratic term in your equation.
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SO... I'm not sure where you need to go with this problem.
I'm going to presume you meant to write x^2: {{{x^2-3x-4=0}}
If that is the case, then this is how you solve it:
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{{{x^2-3x-4=0}}}
Factor equation:
{{{(x-4)(x+1)=0}}}
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Solution set: x={4,-1}
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If I misinterpreted your problem, please re-post it.
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Thanks!