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

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Solution

$x^{2} - (2 b - 1) x + b^{2} - b - 20 = 0$
The roots are:
$x = \frac{(2 b - 1) \pm \sqrt{{(2 b - 1)}^{2} - 4 \times (b^{2} - b - 20)}}{2} = \frac{(2 b - 1) \pm \sqrt{4 b^{2} + 1 - 4 b - 4 b^{2} + 4 b + 80}}{2} = \frac{(2 b - 1) \pm \sqrt{81}}{2} = \frac{(2 b - 1) \pm 9}{2}$
$∴ x = \frac{2 b - 1 + 9}{2} = b + 4 o r, x = \frac{2 b - 1 - 9}{2} = b - 5$

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

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x2−(2b−1)x+(b2−b−20)=0.

x2−(2b−1)x+b2−b−20=0The roots are:x=(2b−1)±√(2b−1)2−4×(b2−b−20)2=(2b−1)±√4b2+1−4b−4b2+4b+802=(2b−1)±√812=(2b−1)±92∴x=2b−1+92=b+4or,x=2b−1−92=b−5

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