What are the solutions to the equation 4y^2+12y-10=0 ?

The solutions to the equation 4y^2+12y-10=0 are y=(-3+sqrt(19))/2 ,y=-(3+sqrt(19))/2

Find the zeros of 4y^2+12y-10=0

The zeros of 4y^2+12y-10=0 are y=(-3+sqrt(19))/2 ,y=-(3+sqrt(19))/2

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4 y^{2} + 12 y - 10 = 0

y = \frac{- 3 + 19}{2}, y = - \frac{3 + 19}{2}

Solution steps

4 y^{2} + 12 y - 10 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 ( - 10 )}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 (- 10) = 419

y_{1, 2} = \frac{- 12 \pm 4 19}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- 12 + 4 19}{2 \cdot 4}, y_{2} = \frac{- 12 - 4 19}{2 \cdot 4}

y = \frac{- 12 + 4 19}{2 \cdot 4} : \frac{- 3 + 19}{2}

y = \frac{- 12 - 4 19}{2 \cdot 4} : - \frac{3 + 19}{2}

The solutions to the quadratic equation are:

y = \frac{- 3 + 19}{2}, y = - \frac{3 + 19}{2}

3 (x + 1) = 3 x + 2

x^{2} - 2 x - 15 \leq 0

a^{3 x} - a^{2 x}

4 w^{2} + 24 w = 0

(x + 6) (1 - x) (x - 4) \leq 0

What are the solutions to the equation 4y^2+12y-10=0 ?
The solutions to the equation 4y^2+12y-10=0 are y=(-3+sqrt(19))/2 ,y=-(3+sqrt(19))/2
Find the zeros of 4y^2+12y-10=0
The zeros of 4y^2+12y-10=0 are y=(-3+sqrt(19))/2 ,y=-(3+sqrt(19))/2