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

The solutions to the equation b^2+b=2 are b=1,b=-2

Find the zeros of b^2+b=2

The zeros of b^2+b=2 are b=1,b=-2

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b^{2} + b = 2

b = 1, b = - 2

Solution steps

b^{2} + b = 2

Move

2

to the left side

b^{2} + b - 2 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 2) = 3

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

b = 1, b = - 2

x^{2} + 11 x

y^{2} - 4 y - 5 = 0

x^{2} + 16 x + 24 = 0

∣ x ∣ = 2

\frac{9}{100}

What are the solutions to the equation b^2+b=2 ?
The solutions to the equation b^2+b=2 are b=1,b=-2
Find the zeros of b^2+b=2
The zeros of b^2+b=2 are b=1,b=-2