Question 13991
Okay, with this equation it's all about factoring. To factor a quadratic that hasn't got a coefficient in front of x^2, we basically just ask ourselves: what two numbers added together give us (-11) and give us 28 when multiplied. 
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Hmm ... don't (-7) and (-4) add to (-11) and multiply to 28??
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Yes, they do.
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So we get the following:
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(X-7)(X-4) = 0
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We're looking for all instances when y = 0. Therefore, we can deduct that:
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X = 7, and X = 4
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As (7-7)(7-4) = 0 >> (0)(3) = 0 >> 0 = 0
And (4-7)(4-4) = 0 >> (-3)(0) = 0 >> 0 = 0