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

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

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

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

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y^{2} - y - 4 = 0

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

Decimal Notation

y = 2.56155 \dots, y = - 1.56155 \dots

Solution steps

y^{2} - y - 4 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x^{2} + 12 x + 36 = 6

5 x^{2} - 4 = 16

8 x^{2} + 22 x + 12 = 0

- 3 x^{2} + 4 x + 59 = - 8 x - 4

(x - 5) (40 - x) = 4 (40 - x)

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