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}

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}

\frac{0}{8}

v^{2} = 1

32 + 42

3 k + 5 = 23

\frac{- 12}{3}

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