Let's solve this with the quadratic formula, which is: [tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Let's plug in the values:
[tex]\frac{-0+\sqrt{0^2-4\cdot \:5\left(-125\right)}}{2\cdot \:5}[/tex]
Remove exponents:
[tex]=\frac{-0+\sqrt{0-\left(-125\right)\cdot \:4\cdot \:5}}{2\cdot \:5}[/tex]
Simplify.
[tex]=\frac{-0+\sqrt{0-\left(-125\right)\cdot \:4\cdot \:5}}{10}[/tex]
[tex]=\frac{\sqrt{-\left(-125\right)\cdot \:20}}{10}[/tex]
[tex]=\frac{50}{10}[/tex]
[tex]=\pm 5[/tex]
This leaves us with the final solutions being:
[tex]x=5,\:x=-5[/tex]
Let me know if you have any questions regarding this problem!
Thanks!
-TetraFish
