What are the solutions to the equation y(y+1)=4 ?

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

Find the zeros of y(y+1)=4

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

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y (y + 1) = 4

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

Decimal Notation

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

Solution steps

y (y + 1) = 4

Expand

y (y + 1) : y^{2} + y

y^{2} + y = 4

Move

4

to the left side

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}

\frac{1}{4} x^{2} + 4 x + 2 = 0

5 m^{2} - 13 m - 6 = 0

x, x^{2} + y^{2} - x - 5 y + 18 = 0

5 x^{2} + 7 x - 9 = 4 x^{2} + x - 18

- 3 x^{2} + 8 x = 7

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