Answer: x = 10 is the positive root
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Work Shown:
First get everything to one side
![x^2 + 5x = 150\\\\x^2 + 5x - 150 = 0\\\\](https://img.qammunity.org/2021/formulas/mathematics/high-school/286vywriwscz7absvur977vinarba9g2j3.png)
Now use the quadratic formula
Plug in a = 1, b = 5, c = -150.
![x = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(5)\pm√((5)^2-4(1)(-150)))/(2(1))\\\\x = (-5\pm√(625))/(2)\\\\x = (-5\pm25)/(2)\\\\x = (-5+25)/(2) \ \text{ or } \ x = (-5-25)/(2)\\\\x = (20)/(2) \ \text{ or } \ x = (-30)/(2)\\\\x = 10 \ \text{ or } \ x = -15\\\\](https://img.qammunity.org/2021/formulas/mathematics/high-school/bcnkncci935ulo3nro4aizdkf17yfktr9k.png)
Factoring, completing the square, or graphing are alternative methods to get these two answers. We see that the positive root is x = 10.