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

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

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

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

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

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

Decimal Notation

y = 1.5, y = - 1

Solution steps

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 2 (- 3) = 5

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

Separate the solutions

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

y = \frac{- ( - 1 ) + 5}{2 \cdot 2} : \frac{3}{2}

y = \frac{- ( - 1 ) - 5}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

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

- x^{2} + 600 x - 90000 = 0

x, 12 x^{2} - 144 x + 299 = 0

x^{2} = a + b

2 x (x + 9) = x (x - 2)

x^{2} + 8 x + 15 = 0

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