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

The solutions to the equation 2y^2+5y-3=0 are y= 1/2 ,y=-3

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

The zeros of 2y^2+5y-3=0 are y= 1/2 ,y=-3

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

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

Decimal Notation

y = 0.5, y = - 3

Solution steps

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

Solve with the quadratic formula

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

5^{2} - 4 \cdot 2 (- 3) = 7

y_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 5 + 7}{2 \cdot 2}, y_{2} = \frac{- 5 - 7}{2 \cdot 2}

y = \frac{- 5 + 7}{2 \cdot 2} : \frac{1}{2}

y = \frac{- 5 - 7}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

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

2 x (w - 3) - 3 y (w - 3)

∣ x + 2 ∣ \geq 5

7 (x - 3) = 4 (x + 5)

29 = 5^{2} + x^{2}

3 k^{3} + k^{2} - 4 k

What are the solutions to the equation 2y^2+5y-3=0 ?
The solutions to the equation 2y^2+5y-3=0 are y= 1/2 ,y=-3
Find the zeros of 2y^2+5y-3=0
The zeros of 2y^2+5y-3=0 are y= 1/2 ,y=-3