How do you solve $x (x - 4) - 2 = 43$ $x(x-4)-2=43$ ?

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How do you solve $x (x - 4) - 2 = 43$ $x(x-4)-2=43$ ?

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Answer 1 · 2015-03-27T17:49:55

I would start by multiplying the $x$ $x$ to get rid of the bracket:
$x^{2} - 4 x - 2 = 43$ $x^2-4x-2=43$
rearranging:
$x^{2} - 4 x - 2 - 43 = 0$ $x^2-4x-2-43=0$
$x^{2} - 4 x - 45 = 0$ $x^2-4x-45=0$
so that:
$x_{1, 2} = \frac{4 \pm \sqrt{16 - 4 (1 \cdot - 45)}}{2} = \frac{4 \pm \sqrt{196}}{2} = \frac{4 \pm 14}{2}$ $x_(1,2)=(4+-sqrt(16-4(1*-45)))/2=(4+-sqrt(196))/2=(4+-14)/2$
So, you get two solutions:
$x_{1} = \frac{18}{2} = 9$ $x_1=18/2=9$
$x_{2} = - \frac{10}{2} = - 5$ $x_2=-10/2=-5$

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How do you solve $x (x - 4) - 2 = 43$ $x(x-4)-2=43$ ?

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