What are the solutions to the equation b(b-5)-9=0 ?

The solutions to the equation b(b-5)-9=0 are b=(5+sqrt(61))/2 ,b=(5-sqrt(61))/2

Find the zeros of b(b-5)-9=0

The zeros of b(b-5)-9=0 are b=(5+sqrt(61))/2 ,b=(5-sqrt(61))/2

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b (b - 5) - 9 = 0

b = \frac{5 + 61}{2}, b = \frac{5 - 61}{2}

Decimal Notation

b = 6.40512 \dots, b = - 1.40512 \dots

Solution steps

b (b - 5) - 9 = 0

Expand

b (b - 5) - 9 : b^{2} - 5 b - 9

b^{2} - 5 b - 9 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 9) = 61

b_{1, 2} = \frac{- ( - 5 ) \pm 61}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- ( - 5 ) + 61}{2 \cdot 1}, b_{2} = \frac{- ( - 5 ) - 61}{2 \cdot 1}

b = \frac{- ( - 5 ) + 61}{2 \cdot 1} : \frac{5 + 61}{2}

b = \frac{- ( - 5 ) - 61}{2 \cdot 1} : \frac{5 - 61}{2}

The solutions to the quadratic equation are:

b = \frac{5 + 61}{2}, b = \frac{5 - 61}{2}

x (x + 6) = 3

2 x^{2} - 1 = x

(x - 1) (2 x - 1) = (x + 1)^{2}

n = x^{2} - 2 x

3 x^{2} - 32 x + 60 = 0

What are the solutions to the equation b(b-5)-9=0 ?
The solutions to the equation b(b-5)-9=0 are b=(5+sqrt(61))/2 ,b=(5-sqrt(61))/2
Find the zeros of b(b-5)-9=0
The zeros of b(b-5)-9=0 are b=(5+sqrt(61))/2 ,b=(5-sqrt(61))/2