What are the solutions to the equation b^2-4b-21=0 ?

The solutions to the equation b^2-4b-21=0 are b=7,b=-3

Find the zeros of b^2-4b-21=0

The zeros of b^2-4b-21=0 are b=7,b=-3

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b^{2} - 4 b - 21 = 0

b = 7, b = - 3

Solution steps

b^{2} - 4 b - 21 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 21 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 21) = 10

b_{1, 2} = \frac{- ( - 4 ) \pm 10}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- ( - 4 ) + 10}{2 \cdot 1}, b_{2} = \frac{- ( - 4 ) - 10}{2 \cdot 1}

b = \frac{- ( - 4 ) + 10}{2 \cdot 1} : 7

b = \frac{- ( - 4 ) - 10}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

b = 7, b = - 3

(\frac{4 x ^{- 2} y ^{3}}{x y ^{- 4}})^{- 2}

a^{2} + 14 a - 51 = 0

(\frac{x}{4})^{2}

(- \frac{1}{4})^{- 3}

21 6^{\frac{1}{3}}

What are the solutions to the equation b^2-4b-21=0 ?
The solutions to the equation b^2-4b-21=0 are b=7,b=-3
Find the zeros of b^2-4b-21=0
The zeros of b^2-4b-21=0 are b=7,b=-3