What are the solutions to the equation 2y^2+y-15=0 ?

The solutions to the equation 2y^2+y-15=0 are y= 5/2 ,y=-3

Find the zeros of 2y^2+y-15=0

The zeros of 2y^2+y-15=0 are y= 5/2 ,y=-3

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2 y^{2} + y - 15 = 0

y = \frac{5}{2}, y = - 3

Decimal Notation

y = 2.5, y = - 3

Solution steps

2 y^{2} + y - 15 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 15 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 15) = 11

y_{1, 2} = \frac{- 1 \pm 11}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 1 + 11}{2 \cdot 2}, y_{2} = \frac{- 1 - 11}{2 \cdot 2}

y = \frac{- 1 + 11}{2 \cdot 2} : \frac{5}{2}

y = \frac{- 1 - 11}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

y = \frac{5}{2}, y = - 3

x^{2} - 9 x

(x - 2)^{2} + (x - 9)^{2} = x^{2}

3 (x - 5)^{2} - 81 = 0

3 x^{2} - 4 x - 20 = 0

x^{2} - 2 x = 0

What are the solutions to the equation 2y^2+y-15=0 ?
The solutions to the equation 2y^2+y-15=0 are y= 5/2 ,y=-3
Find the zeros of 2y^2+y-15=0
The zeros of 2y^2+y-15=0 are y= 5/2 ,y=-3