Question

1. $-3b^2 + 3b + 90 = 0$

Explanation:

Step1: Simplify the equation

Divide all terms by $-3$:
$$\frac{-3b^2}{-3} + \frac{3b}{-3} + \frac{90}{-3} = \frac{0}{-3}$$
$$b^2 - b - 30 = 0$$

Step2: Factor the quadratic

Find two numbers that multiply to $-30$ and add to $-1$ (these are $-6$ and $5$):
$$(b - 6)(b + 5) = 0$$

Step3: Solve for $b$

Set each factor equal to 0:
$b - 6 = 0$ or $b + 5 = 0$
$b = 6$ or $b = -5$

Answer:

$b = 6$ or $b = -5$