What are the solutions to the equation b^2+6b=41 ?

The solutions to the equation b^2+6b=41 are b=-3+5sqrt(2),b=-3-5sqrt(2)

Find the zeros of b^2+6b=41

The zeros of b^2+6b=41 are b=-3+5sqrt(2),b=-3-5sqrt(2)

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b^{2} + 6 b = 41

b = - 3 + 52, b = - 3 - 52

Decimal Notation

b = 4.07106 \dots, b = - 10.07106 \dots

Solution steps

b^{2} + 6 b = 41

Move

41

to the left side

b^{2} + 6 b - 41 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot ( - 41 )}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot (- 41) = 102

b_{1, 2} = \frac{- 6 \pm 10 2}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- 6 + 10 2}{2 \cdot 1}, b_{2} = \frac{- 6 - 10 2}{2 \cdot 1}

b = \frac{- 6 + 10 2}{2 \cdot 1} : - 3 + 52

b = \frac{- 6 - 10 2}{2 \cdot 1} : - 3 - 52

The solutions to the quadratic equation are:

b = - 3 + 52, b = - 3 - 52

- 5 x^{2} + 3 x + 10 = 0

1500 x^{2} + 200 x + 18 = 0

x, y x^{2} - 1 + 6 y = 5 x y + x

x^{2} + 3 x - 27 = 0

250 = (4 w - 2) (w)

What are the solutions to the equation b^2+6b=41 ?
The solutions to the equation b^2+6b=41 are b=-3+5sqrt(2),b=-3-5sqrt(2)
Find the zeros of b^2+6b=41
The zeros of b^2+6b=41 are b=-3+5sqrt(2),b=-3-5sqrt(2)