What are the solutions to the equation y^2+22y-4=0 ?

The solutions to the equation y^2+22y-4=0 are y=-11+5sqrt(5),y=-11-5sqrt(5)

Find the zeros of y^2+22y-4=0

The zeros of y^2+22y-4=0 are y=-11+5sqrt(5),y=-11-5sqrt(5)

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y^{2} + 22 y - 4 = 0

y = - 11 + 55, y = - 11 - 55

Decimal Notation

y = 0.18033 \dots, y = - 22.18033 \dots

Solution steps

y^{2} + 22 y - 4 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 22 \pm 2 2 ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

2 2^{2} - 4 \cdot 1 \cdot (- 4) = 105

y_{1, 2} = \frac{- 22 \pm 10 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 22 + 10 5}{2 \cdot 1}, y_{2} = \frac{- 22 - 10 5}{2 \cdot 1}

y = \frac{- 22 + 10 5}{2 \cdot 1} : - 11 + 55

y = \frac{- 22 - 10 5}{2 \cdot 1} : - 11 - 55

The solutions to the quadratic equation are:

y = - 11 + 55, y = - 11 - 55

5^{2}

\frac{7 n}{n + 1} + \frac{8}{n - 7}

27 y^{3} + 8

10 x^{2} + 11 x - 6

2 - - 2

What are the solutions to the equation y^2+22y-4=0 ?
The solutions to the equation y^2+22y-4=0 are y=-11+5sqrt(5),y=-11-5sqrt(5)
Find the zeros of y^2+22y-4=0
The zeros of y^2+22y-4=0 are y=-11+5sqrt(5),y=-11-5sqrt(5)