What are the solutions to the equation 2y^2+6y-7=0 ?

The solutions to the equation 2y^2+6y-7=0 are y=(-3+sqrt(23))/2 ,y=-(3+sqrt(23))/2

Find the zeros of 2y^2+6y-7=0

The zeros of 2y^2+6y-7=0 are y=(-3+sqrt(23))/2 ,y=-(3+sqrt(23))/2

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2 y^{2} + 6 y - 7 = 0

y = \frac{- 3 + 23}{2}, y = - \frac{3 + 23}{2}

Decimal Notation

y = 0.89791 \dots, y = - 3.89791 \dots

Solution steps

2 y^{2} + 6 y - 7 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

6^{2} - 4 \cdot 2 (- 7) = 223

y_{1, 2} = \frac{- 6 \pm 2 23}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 6 + 2 23}{2 \cdot 2}, y_{2} = \frac{- 6 - 2 23}{2 \cdot 2}

y = \frac{- 6 + 2 23}{2 \cdot 2} : \frac{- 3 + 23}{2}

y = \frac{- 6 - 2 23}{2 \cdot 2} : - \frac{3 + 23}{2}

The solutions to the quadratic equation are:

y = \frac{- 3 + 23}{2}, y = - \frac{3 + 23}{2}

p, p^{2} = a^{3}

t, v = t^{2} - 38

5.6 - 7 - 5 x - 19.3 x^{2} = 0

x^{2} + x + 2 = 9

3 t^{2} - 8 t = 0

What are the solutions to the equation 2y^2+6y-7=0 ?
The solutions to the equation 2y^2+6y-7=0 are y=(-3+sqrt(23))/2 ,y=-(3+sqrt(23))/2
Find the zeros of 2y^2+6y-7=0
The zeros of 2y^2+6y-7=0 are y=(-3+sqrt(23))/2 ,y=-(3+sqrt(23))/2