Find the roots of each of the following equations, if they exist, by applying the quadratic formula:
$x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0$

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Solution

$x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0$

Solving the Q.E in x, we have

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2}$

Discriminant, $D = b^{2} - 4 a c$

$D = (2 b - 1)^{2} - 4 * 1 * (b^{2} - b - 20)$

$= (4 b^{2} + 1 - 4 b - 4 b^{2} + 4 b + 80) = 81 > 0, so solution exists$

$x = \frac{- b \pm \sqrt{D}}{2}$

$x = \frac{(2 b - 1) \pm \sqrt{81}}{2}$

$x = \frac{2 b - 1 \pm 9}{2}$

$x = \frac{2 b - 1 + 9}{2} o r \frac{2 b - 1 - 9}{2}$

$x = \frac{2 b + 8}{2} o r \frac{2 b - 10}{2}$

$x = b + 4 o r b - 5$

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