Question
Solve the following quadratic equation: $x^2 - (2b - 1)x + (b^2 - b - 20) = 0$

Answer

$x^2 - (2b - 1)x + (b^2 - b - 20) = 0$
$\Rightarrow x^2 - (2b - 1)x + (b^2 - 5b + 4b - 20) = 0$
$\Rightarrow x^2 - (2b - 1)x + [b(b - 5) + 4(b - 5)] = 0$
$\Rightarrow x^2- (2b - 1)x + (b - 5)(b + 4) = 0$
$\Rightarrow x^2 - (2b - 1)x - (b + 4)x + (b - 5)(b + 4) = 0$
$\Rightarrow x[x - (b - 5)] - (b + 4)[x - (b - 5)] = 0$
$\Rightarrow [x - (b - 5)][x - (b + 4)] = 0$
$\Rightarrow x - (b - 5) = 0$ or $x - (b + 4) = 0$
$\Rightarrow x = b - 5$ or $x = b + 4$

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