What are the solutions to the equation b^2+9b+17=7 ?

The solutions to the equation b^2+9b+17=7 are b=(-9+sqrt(41))/2 ,b=(-9-sqrt(41))/2

Find the zeros of b^2+9b+17=7

The zeros of b^2+9b+17=7 are b=(-9+sqrt(41))/2 ,b=(-9-sqrt(41))/2

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b^{2} + 9 b + 17 = 7

b = \frac{- 9 + 41}{2}, b = \frac{- 9 - 41}{2}

Decimal Notation

b = - 1.29843 \dots, b = - 7.70156 \dots

Solution steps

b^{2} + 9 b + 17 = 7

Move

7

to the left side

b^{2} + 9 b + 10 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

9^{2} - 4 \cdot 1 \cdot 10 = 41

b_{1, 2} = \frac{- 9 \pm 41}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- 9 + 41}{2 \cdot 1}, b_{2} = \frac{- 9 - 41}{2 \cdot 1}

b = \frac{- 9 + 41}{2 \cdot 1} : \frac{- 9 + 41}{2}

b = \frac{- 9 - 41}{2 \cdot 1} : \frac{- 9 - 41}{2}

The solutions to the quadratic equation are:

b = \frac{- 9 + 41}{2}, b = \frac{- 9 - 41}{2}

x^{2} = - 32

8 - ∣ 2 x - 3 ∣ = 5

(x^{- 5} y^{\frac{1}{3}})^{- \frac{3}{5}}

2 (3 w + 8) = 40

x^{2} + 16 x + 15

What are the solutions to the equation b^2+9b+17=7 ?
The solutions to the equation b^2+9b+17=7 are b=(-9+sqrt(41))/2 ,b=(-9-sqrt(41))/2
Find the zeros of b^2+9b+17=7
The zeros of b^2+9b+17=7 are b=(-9+sqrt(41))/2 ,b=(-9-sqrt(41))/2