What are the solutions to the equation 2y^2+10y=-7 ?

The solutions to the equation 2y^2+10y=-7 are y=(-5+sqrt(11))/2 ,y=-(5+sqrt(11))/2

Find the zeros of 2y^2+10y=-7

The zeros of 2y^2+10y=-7 are y=(-5+sqrt(11))/2 ,y=-(5+sqrt(11))/2

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2 y^{2} + 10 y = - 7

y = \frac{- 5 + 11}{2}, y = - \frac{5 + 11}{2}

Decimal Notation

y = - 0.84168 \dots, y = - 4.15831 \dots

Solution steps

2 y^{2} + 10 y = - 7

Move

7

to the left side

2 y^{2} + 10 y + 7 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

1 0^{2} - 4 \cdot 2 \cdot 7 = 211

y_{1, 2} = \frac{- 10 \pm 2 11}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 10 + 2 11}{2 \cdot 2}, y_{2} = \frac{- 10 - 2 11}{2 \cdot 2}

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

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

The solutions to the quadratic equation are:

y = \frac{- 5 + 11}{2}, y = - \frac{5 + 11}{2}

(x - 1) (x - 2) (x - 6) > 0

9 x^{2} - 9 x + 2

2 (x - 3) = (x - 1) + 7

\frac{1 \cdot 1 0 ^{- 14}}{4.4 \cdot 10 - 4}

- \frac{1}{2} - 4

What are the solutions to the equation 2y^2+10y=-7 ?
The solutions to the equation 2y^2+10y=-7 are y=(-5+sqrt(11))/2 ,y=-(5+sqrt(11))/2
Find the zeros of 2y^2+10y=-7
The zeros of 2y^2+10y=-7 are y=(-5+sqrt(11))/2 ,y=-(5+sqrt(11))/2