What are the solutions to the equation b^2+9b+7=-6 ?

The solutions to the equation b^2+9b+7=-6 are b=(-9+sqrt(29))/2 ,b=(-9-sqrt(29))/2

Find the zeros of b^2+9b+7=-6

The zeros of b^2+9b+7=-6 are b=(-9+sqrt(29))/2 ,b=(-9-sqrt(29))/2

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b^{2} + 9 b + 7 = - 6

b = \frac{- 9 + 29}{2}, b = \frac{- 9 - 29}{2}

Decimal Notation

b = - 1.80741 \dots, b = - 7.19258 \dots

Solution steps

b^{2} + 9 b + 7 = - 6

Move

6

to the left side

b^{2} + 9 b + 13 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

9^{2} - 4 \cdot 1 \cdot 13 = 29

b_{1, 2} = \frac{- 9 \pm 29}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- 9 + 29}{2 \cdot 1}, b_{2} = \frac{- 9 - 29}{2 \cdot 1}

b = \frac{- 9 + 29}{2 \cdot 1} : \frac{- 9 + 29}{2}

b = \frac{- 9 - 29}{2 \cdot 1} : \frac{- 9 - 29}{2}

The solutions to the quadratic equation are:

b = \frac{- 9 + 29}{2}, b = \frac{- 9 - 29}{2}

- 10 z = 40

\frac{16}{5} + 1

3 x^{2} + 12 x - (x + 4) = 0

\frac{9 π}{6}

12 y^{4} + 48 y^{5} - 24 y^{6}

What are the solutions to the equation b^2+9b+7=-6 ?
The solutions to the equation b^2+9b+7=-6 are b=(-9+sqrt(29))/2 ,b=(-9-sqrt(29))/2
Find the zeros of b^2+9b+7=-6
The zeros of b^2+9b+7=-6 are b=(-9+sqrt(29))/2 ,b=(-9-sqrt(29))/2