20b-180+20b-9b+b²=0

To solve the equation \(20b - 180 + 20b - 9b + b^2 = 0\), we first combine like terms.

1. Combine the \(b\) terms: \[ 20b + 20b - 9b = (20b + 20b - 9b) = 31b. \]

So the equation simplifies to: \[ b^2 + 31b - 180 = 0. \]

Next, we can use the quadratic formula to find the values of \(b\). The quadratic formula is given by: \[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}, \] where \(A = 1\), \(B = 31\), and \(C = -180\).

2. Calculate the discriminant: \[ B^2 - 4AC = 31^2 - 4(1)(-180) = 961 + 720 = 1681. \]

3. Now, substitute back into the quadratic formula: \[ b = \frac{-31 \pm \sqrt{1681}}{2(1)} = \frac{-31 \pm 41}{2}. \]

4. Calculate the two possible values for \(b\):

   • For the positive root: \[ b = \frac{-31 + 41}{2} = \frac{10}{2} = 5. \]
   • For the negative root: \[ b = \frac{-31 - 41}{2} = \frac{-72}{2} = -36. \]

Thus, the solutions for the equation are: \[ b = 5 \quad \text{and} \quad b = -36. \]