What are the solutions to the equation b^2-4b-5=0 ?

The solutions to the equation b^2-4b-5=0 are b=5,b=-1

Find the zeros of b^2-4b-5=0

The zeros of b^2-4b-5=0 are b=5,b=-1

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b^{2} - 4 b - 5 = 0

b = 5, b = - 1

Solution steps

b^{2} - 4 b - 5 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 5) = 6

b_{1, 2} = \frac{- ( - 4 ) \pm 6}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- ( - 4 ) + 6}{2 \cdot 1}, b_{2} = \frac{- ( - 4 ) - 6}{2 \cdot 1}

b = \frac{- ( - 4 ) + 6}{2 \cdot 1} : 5

b = \frac{- ( - 4 ) - 6}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

b = 5, b = - 1

72 x^{2} - 120 x + 50

(z + 4)^{2} = 3

(64 v^{3})^{\frac{4}{3}}

a^{4} - 256

x^{9} - 1

What are the solutions to the equation b^2-4b-5=0 ?
The solutions to the equation b^2-4b-5=0 are b=5,b=-1
Find the zeros of b^2-4b-5=0
The zeros of b^2-4b-5=0 are b=5,b=-1