Question

please answer this question ​

Answer 1

$\huge\red {Solution}$

Given Equation :-

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

▶Finding the value of ${x}$ :-

$x = \frac{(2b - 1) + \sqrt{ {(2b - 1) - 4( {b}^{2} - b - 20)} } }{2}$

$x = \frac{(2b - 1) + 9}{2}$

$x = \frac{2b + 8}{2} \: \: and \: \: x = \frac{2b - 10}{2}$

⏩Therefore, the values of ${x}$ are :-

$x = b + 4 \: \: and \: \: x = b - 5$

$\huge\mathcal {It\:Might\:Help\:You}$

I'm in Singh Karishma

and I'm in class 10th

Answer 2

Answer:

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