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solve x^2-3x-4=0
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D = (-3)^2 - 4 * 1 * (-4) = 9 + 16 = 25 x1 = [-(-3) - sqrt(25)] / 2 = [ 3 - 5] / 2 = -2 / 2 = -1
x^2-3x-4=0 (x+1)(x-4)=0
you could always try factoring!
x2 = [-(-3) + sqrt(25)] / 2 = [ 3 + 5] / 2 = 8 / 2 = 4 Sorry the site crashed on me, but yea you could try factoring too, but I never really bothered learning it haha, if the question asks me to factor, I just solve it out and then do (x-x1)(x-x2) to factor which will be (x-(-1))(x-4) = (x+1)(x-4)
I prefer factoring if there are small numbers. My technique is pretty easy: x^2-3x-4=0 we have -3x and -4. -4 is -2*2 or 4*-1 or -4*1 -3x is 1x-4x (1, -4) so we can see that (1,-4) will be the factors
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