What are the solutions to the equation 2y^2-3y-9=0 ?

The solutions to the equation 2y^2-3y-9=0 are y=3,y=-3/2

Find the zeros of 2y^2-3y-9=0

The zeros of 2y^2-3y-9=0 are y=3,y=-3/2

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2 y^{2} - 3 y - 9 = 0

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

Decimal Notation

y = 3, y = - 1.5

Solution steps

2 y^{2} - 3 y - 9 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 9) = 9

y_{1, 2} = \frac{- ( - 3 ) \pm 9}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 9}{2 \cdot 2}, y_{2} = \frac{- ( - 3 ) - 9}{2 \cdot 2}

y = \frac{- ( - 3 ) + 9}{2 \cdot 2} : 3

y = \frac{- ( - 3 ) - 9}{2 \cdot 2} : - \frac{3}{2}

The solutions to the quadratic equation are:

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

x^{2} - 14 x - 11 = - 30

(x + 1)^{2} - 1 \frac{1}{3} = 0

x, (x + 8)^{2} = 16

y, y^{2} + x^{2} = 0

x^{2} - 6 x - 33 = - 4 x + 2

What are the solutions to the equation 2y^2-3y-9=0 ?
The solutions to the equation 2y^2-3y-9=0 are y=3,y=-3/2
Find the zeros of 2y^2-3y-9=0
The zeros of 2y^2-3y-9=0 are y=3,y=-3/2