Answer:
Quadratic equations : ax² +bx +c
b² -6b +5= -4
b² -6b +5 +4=0
b² -6b +9=0
Instead of x, the algebraic term is b.
So the coefficient of b², a, is equal to 1
coefficient of b, b, is equals to -6
and the constant c is 9.
Note that the value of b (coefficient of b term) is -6 not 6.
The formula you wrote is correct however when you substitute the value of b in "-b", it should have been -(-6) so you would get 6 instead.
[tex]b = \frac{6 + \sqrt{( - 6)^{2} - 4(1)(9)}} \: or \: b = \frac{6 - \sqrt{( - 6)^{2} - 4(1)(9)(1)}} \\ \\b = \frac{6 + \sqrt{0} }{2} \: or \: b = \frac{6 - \sqrt{0} }{2} \\ b = \frac{6}{2} \: or \: b = \frac{6}{2} \\ b = 3 \: \: or \: b = 3[/tex]
Thus the answer should be b=3
