Question

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

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

it's a quadratic equation, we can find the value of x by quadratic formula,

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

${= 4b^2 + 1 - 4b - 4b^2 + 4b +80}$

=81

x = 2b-1 +- √81 /2

${x = 2b - 1 + 9 / 2 \:or\: 2b - 1 - 9 /2}$

$\large\underline{x = (b+4) or (b+5)}$

$\huge\mathfrak{brainliest\:plz}$

Answer 2

Answer:

By solving, we get x = (b+4) or (b+5)