What are the solutions to the equation roots x^2-x-4 ?

The solutions to the equation roots x^2-x-4 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2

Find the zeros of roots x^2-x-4

The zeros of roots x^2-x-4 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2

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roots

x^{2} - x - 4

x = \frac{1 + 17}{2}, x = \frac{1 - 17}{2}

Solution steps

x^{2} - x - 4

The roots are the intercepts with the x-axis

(y = 0)

x^{2} - x - 4 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 4) = 17

x_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 17}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 17}{2 \cdot 1}

x = \frac{- ( - 1 ) + 17}{2 \cdot 1} : \frac{1 + 17}{2}

x = \frac{- ( - 1 ) - 17}{2 \cdot 1} : \frac{1 - 17}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 17}{2}, x = \frac{1 - 17}{2}

y, 7 x^{2} - 63 x y + 13 y^{2} - 16 = 0

\frac{1}{2} x^{2} - 5 = 13

1 + \frac{5}{6} x + \frac{1}{6} x^{2} = 0

5 \cdot (- \frac{x ^{2}}{1} + 2 x + 2) = 1

x, 2 x^{2} - 32 x + 128 = 0

What are the solutions to the equation roots x^2-x-4 ?
The solutions to the equation roots x^2-x-4 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2
Find the zeros of roots x^2-x-4
The zeros of roots x^2-x-4 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2