Answer: Choice C) [tex]x = 4\pm2\sqrt{10}[/tex]
This is the same as saying [tex]x = 4+2\sqrt{10} \ \ \text{ or } \ \ x = 4-2\sqrt{10}[/tex]
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Explanation:
First get everything to one side
x^2 + 13 = 8x + 37
x^2 + 13 - 8x - 37 = 0
x^2 - 8x - 24 = 0
Here we have a quadratic in the form ax^2+bx+c = 0
Note how a = 1, b = -8, c = -24
Those values are plugged into the quadratic formula below
[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-8)\pm\sqrt{(-8)^2-4(1)(-24)}}{2(1)}\\\\x = \frac{8\pm\sqrt{160}}{2}\\\\x = \frac{8\pm\sqrt{16*10}}{2}\\\\x = \frac{8\pm\sqrt{16}*\sqrt{10}}{2}\\\\x = \frac{8\pm4*\sqrt{10}}{2}\\\\x = \frac{2(4\pm2\sqrt{10})}{2}\\\\x = 4\pm2\sqrt{10}[/tex]
This shows the answer is choice C.
The plus minus breaks down that equation these two solutions
[tex]x = 4+2\sqrt{10} \ \ \text{ or } \ \ x = 4-2\sqrt{10}[/tex]
