#### Answer

2 real solutions.

#### Work Step by Step

To find the number of solutions in a quadratic formula, we need to find the determinant. The determinant follows these rules:
If D<0: no real solutions
If D=0: 1 real solution
If D>0: 2 real solutions
The determinant is generally calculated by this formula: $ D={b^2-4ac}$.
The general formula for quadratic equations is: $ax^2+bx+c=0$
Therefore, in this equation, $x^2+6x+1=0$, $a=1$, $b=6$ and $c=1$.
$D= b^2-4ac= 6^2-4*1*1=36-4=32>0$
Therefore, this equation has 2 real solutions.