What are the solutions to the equation b^2+b-56=0 ?

The solutions to the equation b^2+b-56=0 are b=7,b=-8

Find the zeros of b^2+b-56=0

The zeros of b^2+b-56=0 are b=7,b=-8

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b^{2} + b - 56 = 0

b = 7, b = - 8

Solution steps

b^{2} + b - 56 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 56 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 56) = 15

b_{1, 2} = \frac{- 1 \pm 15}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- 1 + 15}{2 \cdot 1}, b_{2} = \frac{- 1 - 15}{2 \cdot 1}

b = \frac{- 1 + 15}{2 \cdot 1} : 7

b = \frac{- 1 - 15}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

b = 7, b = - 8

x^{2} - 4 x + 4 = 81

x^{2} - 15 x + 22 = 0

31 x^{2} + 62 x = 0

X^{2} - 6 X + 8 = 0

m^{2} - 4 m + 1 = 0

What are the solutions to the equation b^2+b-56=0 ?
The solutions to the equation b^2+b-56=0 are b=7,b=-8
Find the zeros of b^2+b-56=0
The zeros of b^2+b-56=0 are b=7,b=-8