Answer: x = 10 is the positive root
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Work Shown:
First get everything to one side
[tex]x^2 + 5x = 150\\\\x^2 + 5x - 150 = 0\\\\[/tex]
Now use the quadratic formula
Plug in a = 1, b = 5, c = -150.
[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(5)\pm\sqrt{(5)^2-4(1)(-150)}}{2(1)}\\\\x = \frac{-5\pm\sqrt{625}}{2}\\\\x = \frac{-5\pm25}{2}\\\\x = \frac{-5+25}{2} \ \text{ or } \ x = \frac{-5-25}{2}\\\\x = \frac{20}{2} \ \text{ or } \ x = \frac{-30}{2}\\\\x = 10 \ \text{ or } \ x = -15\\\\[/tex]
Factoring, completing the square, or graphing are alternative methods to get these two answers. We see that the positive root is x = 10.
