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

The solutions to the equation b^2+2b-5=0 are b=-1+sqrt(6),b=-1-sqrt(6)

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

The zeros of b^2+2b-5=0 are b=-1+sqrt(6),b=-1-sqrt(6)

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

b = - 1 + 6, b = - 1 - 6

Decimal Notation

b = 1.44948 \dots, b = - 3.44948 \dots

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 5) = 26

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

Separate the solutions

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

b = \frac{- 2 + 2 6}{2 \cdot 1} : - 1 + 6

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

The solutions to the quadratic equation are:

b = - 1 + 6, b = - 1 - 6

- 7 x - 1 = - 9 x - 17

\frac{5 x + 25}{x - 1} < 0

2 m^{p} n^{r} + 6 m^{p - 1} n^{r + 1} - 12 m^{p + 1} n^{r - 1}

21 < 3 (5 - 2 x) < 39

\frac{4}{5} (20 x + 5) = 16 x - 2

What are the solutions to the equation b^2+2b-5=0 ?
The solutions to the equation b^2+2b-5=0 are b=-1+sqrt(6),b=-1-sqrt(6)
Find the zeros of b^2+2b-5=0
The zeros of b^2+2b-5=0 are b=-1+sqrt(6),b=-1-sqrt(6)