What are the solutions to the equation 4y^2+17y+17=1 ?

The solutions to the equation 4y^2+17y+17=1 are y=(-17+sqrt(33))/8 ,y=(-17-sqrt(33))/8

Find the zeros of 4y^2+17y+17=1

The zeros of 4y^2+17y+17=1 are y=(-17+sqrt(33))/8 ,y=(-17-sqrt(33))/8

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4 y^{2} + 17 y + 17 = 1

y = \frac{- 17 + 33}{8}, y = \frac{- 17 - 33}{8}

Decimal Notation

y = - 1.40692 \dots, y = - 2.84307 \dots

Solution steps

4 y^{2} + 17 y + 17 = 1

Move

1

to the left side

4 y^{2} + 17 y + 16 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 17 \pm 1 7 ^{2} - 4 \cdot 4 \cdot 16}{2 \cdot 4}

1 7^{2} - 4 \cdot 4 \cdot 16 = 33

y_{1, 2} = \frac{- 17 \pm 33}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- 17 + 33}{2 \cdot 4}, y_{2} = \frac{- 17 - 33}{2 \cdot 4}

y = \frac{- 17 + 33}{2 \cdot 4} : \frac{- 17 + 33}{8}

y = \frac{- 17 - 33}{2 \cdot 4} : \frac{- 17 - 33}{8}

The solutions to the quadratic equation are:

y = \frac{- 17 + 33}{8}, y = \frac{- 17 - 33}{8}

(2 x + 5) (x - 1) = 12 x

x^{2} - 4 x - 41 = 4

5 - x^{2} = 3 - x

y^{2} - 22 y + 21 = 0

a, h^{2} = a^{2} + b^{2}

What are the solutions to the equation 4y^2+17y+17=1 ?
The solutions to the equation 4y^2+17y+17=1 are y=(-17+sqrt(33))/8 ,y=(-17-sqrt(33))/8
Find the zeros of 4y^2+17y+17=1
The zeros of 4y^2+17y+17=1 are y=(-17+sqrt(33))/8 ,y=(-17-sqrt(33))/8