how to solve
(b^2+2)+b=77
Sayar:
b^2 + b - 75 = 0 => b = (-1 +/-(1 - 4×-75×1)^0.5 ÷ 2
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(b² + 2)+b = 77
b² + b + 2 = 77
b² + b - 75 = 0
Here
a = 1
b = 1
c = - 75
b = [-b ± √(b² - 4ac)]/2a
b = [-1 ± √(-1)² - 4*1*-75]/2*1
b = [- 1 ± √(1 + 300)]/2
b = (- 1 + √301)/2 or (- 1 - √301)/2
Hope this helps you.
b² + b + 2 = 77
b² + b - 75 = 0
Here
a = 1
b = 1
c = - 75
b = [-b ± √(b² - 4ac)]/2a
b = [-1 ± √(-1)² - 4*1*-75]/2*1
b = [- 1 ± √(1 + 300)]/2
b = (- 1 + √301)/2 or (- 1 - √301)/2
Hope this helps you.
Answered by
14
Hope it's help you..
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