Find all possible values of \( b: x^{2}+b x-36 \)

To find all possible values of \( b \) such that the quadratic equation \( x^{2}+b x-36 \) has real solutions, we need to look at the discriminant of the equation. The discriminant \( D \) is given by: \[ D = b^2 - 4ac \] Here, \( a = 1 \), \( b = b \), and \( c = -36 \). Thus, the discriminant becomes: \[ D = b^2 - 4(1)(-36) = b^2 + 144 \] For the quadratic to have real solutions, we require \( D \geq 0 \): \[ b^2 + 144 \geq 0 \] Since \( b^2 \) is always non-negative, \( b^2 + 144 \) is always positive. Therefore, the quadratic equation \( x^{2}+b x-36 \) will have real solutions for any value of \( b \). In conclusion, all possible values of \( b \) are: \[ \text{All real numbers } b \]