What are the solutions to the equation 2+3b-b^2=3 ?

The solutions to the equation 2+3b-b^2=3 are b=(3-sqrt(5))/2 ,b=(3+sqrt(5))/2

Find the zeros of 2+3b-b^2=3

The zeros of 2+3b-b^2=3 are b=(3-sqrt(5))/2 ,b=(3+sqrt(5))/2

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2 + 3 b - b^{2} = 3

b = \frac{3 - 5}{2}, b = \frac{3 + 5}{2}

Decimal Notation

b = 0.38196 \dots, b = 2.61803 \dots

Solution steps

2 + 3 b - b^{2} = 3

Move

3

to the left side

- b^{2} + 3 b - 1 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) ( - 1 )}{2 ( - 1 )}

3^{2} - 4 (- 1) (- 1) = 5

b_{1, 2} = \frac{- 3 \pm 5}{2 ( - 1 )}

Separate the solutions

b_{1} = \frac{- 3 + 5}{2 ( - 1 )}, b_{2} = \frac{- 3 - 5}{2 ( - 1 )}

b = \frac{- 3 + 5}{2 ( - 1 )} : \frac{3 - 5}{2}

b = \frac{- 3 - 5}{2 ( - 1 )} : \frac{3 + 5}{2}

The solutions to the quadratic equation are:

b = \frac{3 - 5}{2}, b = \frac{3 + 5}{2}

6 m - 2

8 (x + 1) = 7 - 2 (8 x + 8)

- 9 \leq 2 x - 7 < - 1

(z - 7) (z + 2) = 0

9 p^{2} + 12 p + 4

What are the solutions to the equation 2+3b-b^2=3 ?
The solutions to the equation 2+3b-b^2=3 are b=(3-sqrt(5))/2 ,b=(3+sqrt(5))/2
Find the zeros of 2+3b-b^2=3
The zeros of 2+3b-b^2=3 are b=(3-sqrt(5))/2 ,b=(3+sqrt(5))/2