What are the solutions to the equation b^2+16b-49=0 ?

The solutions to the equation b^2+16b-49=0 are b=-8+sqrt(113),b=-8-sqrt(113)

Find the zeros of b^2+16b-49=0

The zeros of b^2+16b-49=0 are b=-8+sqrt(113),b=-8-sqrt(113)

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b^{2} + 16 b - 49 = 0

b = - 8 + 113, b = - 8 - 113

Decimal Notation

b = 2.63014 \dots, b = - 18.63014 \dots

Solution steps

b^{2} + 16 b - 49 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot ( - 49 )}{2 \cdot 1}

1 6^{2} - 4 \cdot 1 \cdot (- 49) = 2113

b_{1, 2} = \frac{- 16 \pm 2 113}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- 16 + 2 113}{2 \cdot 1}, b_{2} = \frac{- 16 - 2 113}{2 \cdot 1}

b = \frac{- 16 + 2 113}{2 \cdot 1} : - 8 + 113

b = \frac{- 16 - 2 113}{2 \cdot 1} : - 8 - 113

The solutions to the quadratic equation are:

b = - 8 + 113, b = - 8 - 113

5 x^{2} + 12 x + 10 = 0

9 x^{2} + b x + 4 = 0

x^{2} + 10 + 24 = 0

x^{(2)} + 6 x + 5 = 0

4 x + 3 = 4 x^{2}

What are the solutions to the equation b^2+16b-49=0 ?
The solutions to the equation b^2+16b-49=0 are b=-8+sqrt(113),b=-8-sqrt(113)
Find the zeros of b^2+16b-49=0
The zeros of b^2+16b-49=0 are b=-8+sqrt(113),b=-8-sqrt(113)