What are the solutions to the equation 0=x^2-x-4 ?

The solutions to the equation 0=x^2-x-4 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2

Find the zeros of 0=x^2-x-4

The zeros of 0=x^2-x-4 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2

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0 = x^{2} - x - 4

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

Decimal Notation

x = 2.56155 \dots, x = - 1.56155 \dots

Solution steps

0 = x^{2} - x - 4

Switch sides

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}

(w + 5)^{2} - 54 = 0

a^{2} + 4 0^{2} = 5 0^{2}

x, f = I n (x^{2} + 9)

20 = 2 + 0.2 x^{2}

(n - 2) (n + 4) = 27

What are the solutions to the equation 0=x^2-x-4 ?
The solutions to the equation 0=x^2-x-4 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2
Find the zeros of 0=x^2-x-4
The zeros of 0=x^2-x-4 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2