What are the solutions to the equation b^2+sqrt(2)b-4=0 ?

The solutions to the equation b^2+sqrt(2)b-4=0 are b=sqrt(2),b=-2sqrt(2)

Find the zeros of b^2+sqrt(2)b-4=0

The zeros of b^2+sqrt(2)b-4=0 are b=sqrt(2),b=-2sqrt(2)

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b^{2} + 2 b - 4 = 0

b = 2, b = - 22

Decimal Notation

b = 1.41421 \dots, b = - 2.82842 \dots

Solution steps

b^{2} + 2 b - 4 = 0

Solve with the quadratic formula

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

(2)^{2} - 4 \cdot 1 \cdot (- 4) = 32

b_{1, 2} = \frac{- 2 \pm 3 2}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- 2 + 3 2}{2 \cdot 1}, b_{2} = \frac{- 2 - 3 2}{2 \cdot 1}

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

b = \frac{- 2 - 3 2}{2 \cdot 1} : - 22

The solutions to the quadratic equation are:

b = 2, b = - 22

x^{2} - 81 = 0

36 q^{2} - 333 q + 81 = 0

s^{2} + 2 s + 10

4 x^{2} + 69 = 28 x

x^{2} - x - 35 = - x + 1

What are the solutions to the equation b^2+sqrt(2)b-4=0 ?
The solutions to the equation b^2+sqrt(2)b-4=0 are b=sqrt(2),b=-2sqrt(2)
Find the zeros of b^2+sqrt(2)b-4=0
The zeros of b^2+sqrt(2)b-4=0 are b=sqrt(2),b=-2sqrt(2)