Answer:
No real root.
Complex roots:
![x = -1 \pm 2i](https://img.qammunity.org/2021/formulas/mathematics/college/4ffcanlcrv1nlurp8fzcuhz8wfx20beu2q.png)
Explanation:
![x^2 + 2x = -5](https://img.qammunity.org/2021/formulas/mathematics/college/a4hfqjx1h6tomlgzcz30lpxtp37e3i4flt.png)
![x^2 + 2x + 5 = 0](https://img.qammunity.org/2021/formulas/mathematics/college/zwusr2g3f3s5dpshnbz50s6640zbvsf70g.png)
There are no two integers whose product is 5 and whose sum is 2, so this trinomial is not factorable. We can use the quadratic formula.
![x = (-b \pm √(b^2 - 4ac))/(2a)](https://img.qammunity.org/2021/formulas/mathematics/high-school/7zbbsy0oieatyffjcul7nyv81wruhlh37s.png)
![x = (-2 \pm √(2^2 - 4(1)(5)))/(2(1))](https://img.qammunity.org/2021/formulas/mathematics/college/87gzjrfh3lg3hz61aqt3i2o6wi0l9w9gmn.png)
![x = (-2 \pm √(4 - 20))/(2)](https://img.qammunity.org/2021/formulas/mathematics/college/9m1tfnbtlmutyjw5wyygsmbgs1ibo85v5z.png)
![x = (-2 \pm √(-16))/(2)](https://img.qammunity.org/2021/formulas/mathematics/college/vxr66aq1shn3pydxapv9r6usmvgvuutlrp.png)
Since we have a square root of a negative number, there are no real roots. If you have learned complex numbers, then we can continue.
![x = (-2 \pm 4i)/(2)](https://img.qammunity.org/2021/formulas/mathematics/college/z8skjwrhvxg83t58w0uwylqpdml2twqw3r.png)
![x = -1 \pm 2i](https://img.qammunity.org/2021/formulas/mathematics/college/4ffcanlcrv1nlurp8fzcuhz8wfx20beu2q.png)