What are the solutions to the equation x^2-x-b=0 ?

The solutions to the equation x^2-x-b=0 are x=(1+sqrt(1+4b))/2 ,x=(1-sqrt(1+4b))/2

Find the zeros of x^2-x-b=0

The zeros of x^2-x-b=0 are x=(1+sqrt(1+4b))/2 ,x=(1-sqrt(1+4b))/2

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x^{2} - x - b = 0

x = \frac{1 + 1 + 4 b}{2}, x = \frac{1 - 1 + 4 b}{2}

Solution steps

x^{2} - x - b = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot (- b) : 1 + 4 b

x_{1, 2} = \frac{- ( - 1 ) \pm 1 + 4 b}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 1 + 4 b}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 1 + 4 b}{2 \cdot 1}

x = \frac{- ( - 1 ) + 1 + 4 b}{2 \cdot 1} : \frac{1 + 1 + 4 b}{2}

x = \frac{- ( - 1 ) - 1 + 4 b}{2 \cdot 1} : \frac{1 - 1 + 4 b}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 1 + 4 b}{2}, x = \frac{1 - 1 + 4 b}{2}

x, 1^{2} + x^{2} = c \cdot p^{2}

x, 5 x^{2} - 7 x y - 3 y^{2} = 0

x, f (x + 1) = (x + 1)^{2} - 1

x, x^{2} + k (4 x + k - 1) + 2 = 0

150 = 0.0216 d^{2} + 0.0216 (\frac{d + 8}{16})^{2}

What are the solutions to the equation x^2-x-b=0 ?
The solutions to the equation x^2-x-b=0 are x=(1+sqrt(1+4b))/2 ,x=(1-sqrt(1+4b))/2
Find the zeros of x^2-x-b=0
The zeros of x^2-x-b=0 are x=(1+sqrt(1+4b))/2 ,x=(1-sqrt(1+4b))/2