What are the solutions to the equation 5y^2+4y-3=0 ?

The solutions to the equation 5y^2+4y-3=0 are y=(-2+sqrt(19))/5 ,y=-(2+sqrt(19))/5

Find the zeros of 5y^2+4y-3=0

The zeros of 5y^2+4y-3=0 are y=(-2+sqrt(19))/5 ,y=-(2+sqrt(19))/5

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5 y^{2} + 4 y - 3 = 0

y = \frac{- 2 + 19}{5}, y = - \frac{2 + 19}{5}

Decimal Notation

y = 0.47177 \dots, y = - 1.27177 \dots

Solution steps

5 y^{2} + 4 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

4^{2} - 4 \cdot 5 (- 3) = 219

y_{1, 2} = \frac{- 4 \pm 2 19}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- 4 + 2 19}{2 \cdot 5}, y_{2} = \frac{- 4 - 2 19}{2 \cdot 5}

y = \frac{- 4 + 2 19}{2 \cdot 5} : \frac{- 2 + 19}{5}

y = \frac{- 4 - 2 19}{2 \cdot 5} : - \frac{2 + 19}{5}

The solutions to the quadratic equation are:

y = \frac{- 2 + 19}{5}, y = - \frac{2 + 19}{5}

28 x^{2} + 12 x

lo g_{4} (64) = x

- 2 x - 5 = 2 - 4 x - (x - 1)

\frac{10}{12}

\frac{1 9 ^{7} \cdot 1 9 ^{- 5}}{1 9 ^{4}}

What are the solutions to the equation 5y^2+4y-3=0 ?
The solutions to the equation 5y^2+4y-3=0 are y=(-2+sqrt(19))/5 ,y=-(2+sqrt(19))/5
Find the zeros of 5y^2+4y-3=0
The zeros of 5y^2+4y-3=0 are y=(-2+sqrt(19))/5 ,y=-(2+sqrt(19))/5