What are the solutions to the equation 2y^2+14y=-19 ?

The solutions to the equation 2y^2+14y=-19 are y=(-7+sqrt(11))/2 ,y=-(7+sqrt(11))/2

Find the zeros of 2y^2+14y=-19

The zeros of 2y^2+14y=-19 are y=(-7+sqrt(11))/2 ,y=-(7+sqrt(11))/2

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2 y^{2} + 14 y = - 19

y = \frac{- 7 + 11}{2}, y = - \frac{7 + 11}{2}

Decimal Notation

y = - 1.84168 \dots, y = - 5.15831 \dots

Solution steps

2 y^{2} + 14 y = - 19

Move

19

to the left side

2 y^{2} + 14 y + 19 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 2 \cdot 19}{2 \cdot 2}

1 4^{2} - 4 \cdot 2 \cdot 19 = 211

y_{1, 2} = \frac{- 14 \pm 2 11}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 14 + 2 11}{2 \cdot 2}, y_{2} = \frac{- 14 - 2 11}{2 \cdot 2}

y = \frac{- 14 + 2 11}{2 \cdot 2} : \frac{- 7 + 11}{2}

y = \frac{- 14 - 2 11}{2 \cdot 2} : - \frac{7 + 11}{2}

The solutions to the quadratic equation are:

y = \frac{- 7 + 11}{2}, y = - \frac{7 + 11}{2}

4 x + 8 = 20

(- 4)^{- 6} (- 4)^{7}

\frac{3}{4} t = \frac{2}{3}

f + 20 = - 60

x (x - 2)^{2} (x + 1) < 0

What are the solutions to the equation 2y^2+14y=-19 ?
The solutions to the equation 2y^2+14y=-19 are y=(-7+sqrt(11))/2 ,y=-(7+sqrt(11))/2
Find the zeros of 2y^2+14y=-19
The zeros of 2y^2+14y=-19 are y=(-7+sqrt(11))/2 ,y=-(7+sqrt(11))/2