What are the solutions to the equation 8y^2+2y-3=0 ?

The solutions to the equation 8y^2+2y-3=0 are y= 1/2 ,y=-3/4

Find the zeros of 8y^2+2y-3=0

The zeros of 8y^2+2y-3=0 are y= 1/2 ,y=-3/4

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8 y^{2} + 2 y - 3 = 0

y = \frac{1}{2}, y = - \frac{3}{4}

Solution steps

8 y^{2} + 2 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

2^{2} - 4 \cdot 8 (- 3) = 10

y_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 8}

Separate the solutions

y_{1} = \frac{- 2 + 10}{2 \cdot 8}, y_{2} = \frac{- 2 - 10}{2 \cdot 8}

y = \frac{- 2 + 10}{2 \cdot 8} : \frac{1}{2}

y = \frac{- 2 - 10}{2 \cdot 8} : - \frac{3}{4}

The solutions to the quadratic equation are:

y = \frac{1}{2}, y = - \frac{3}{4}

\frac{x + 3}{4} > \frac{2 x - 4}{3}

6 + 9 z + z^{2} = 0

0 = 3 x + 4

4 x^{2} + 9 x^{4}

x^{2} - 2 x > 0

What are the solutions to the equation 8y^2+2y-3=0 ?
The solutions to the equation 8y^2+2y-3=0 are y= 1/2 ,y=-3/4
Find the zeros of 8y^2+2y-3=0
The zeros of 8y^2+2y-3=0 are y= 1/2 ,y=-3/4