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

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

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

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

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x^{2} - x - 4 = 0

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

Decimal Notation

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

Solution steps

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}

(5 w - 8) (4 - w) = 0

y^{\frac{1}{2}} \cdot y^{\frac{1}{2}}

9 m + 3 > - 69 and 2 m - 3 < 17

x^{4} - 13 x^{2} + 40

6 x^{3} - 19 x^{2} + 16 x - 4

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