What are the solutions to the equation 13y^2+10y-3=5y^2 ?

The solutions to the equation 13y^2+10y-3=5y^2 are y= 1/4 ,y=-3/2

Find the zeros of 13y^2+10y-3=5y^2

The zeros of 13y^2+10y-3=5y^2 are y= 1/4 ,y=-3/2

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13 y^{2} + 10 y - 3 = 5 y^{2}

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

Solution steps

13 y^{2} + 10 y - 3 = 5 y^{2}

Move

5 y^{2}

to the left side

8 y^{2} + 10 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

1 0^{2} - 4 \cdot 8 (- 3) = 14

y_{1, 2} = \frac{- 10 \pm 14}{2 \cdot 8}

Separate the solutions

y_{1} = \frac{- 10 + 14}{2 \cdot 8}, y_{2} = \frac{- 10 - 14}{2 \cdot 8}

y = \frac{- 10 + 14}{2 \cdot 8} : \frac{1}{4}

y = \frac{- 10 - 14}{2 \cdot 8} : - \frac{3}{2}

The solutions to the quadratic equation are:

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

2 x + 3 > 0

c^{3} - d^{3}

- 9 x - 23 = 2 x - 1

(\frac{2}{x})^{- 1}

5 x + 2 = - 4

What are the solutions to the equation 13y^2+10y-3=5y^2 ?
The solutions to the equation 13y^2+10y-3=5y^2 are y= 1/4 ,y=-3/2
Find the zeros of 13y^2+10y-3=5y^2
The zeros of 13y^2+10y-3=5y^2 are y= 1/4 ,y=-3/2