What are the solutions to the equation b^2-2b-15=0 ?

The solutions to the equation b^2-2b-15=0 are b=5,b=-3

Find the zeros of b^2-2b-15=0

The zeros of b^2-2b-15=0 are b=5,b=-3

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b^{2} - 2 b - 15 = 0

b = 5, b = - 3

Solution steps

b^{2} - 2 b - 15 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 15) = 8

b_{1, 2} = \frac{- ( - 2 ) \pm 8}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- ( - 2 ) + 8}{2 \cdot 1}, b_{2} = \frac{- ( - 2 ) - 8}{2 \cdot 1}

b = \frac{- ( - 2 ) + 8}{2 \cdot 1} : 5

b = \frac{- ( - 2 ) - 8}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

b = 5, b = - 3

2 x (x + 3) = 0

4 n^{2} = 81

6 x^{2} - 4 x + 5 = 0

t, s = \frac{1}{2} g t^{2}

25 - 70 x + 49 x^{2} = 0

What are the solutions to the equation b^2-2b-15=0 ?
The solutions to the equation b^2-2b-15=0 are b=5,b=-3
Find the zeros of b^2-2b-15=0
The zeros of b^2-2b-15=0 are b=5,b=-3