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

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

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

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

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

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

Decimal Notation

y = 2, y = - 0.5

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 2}, y_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 2}

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

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

The solutions to the quadratic equation are:

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

x^{2} + x - 99 = 0

p^{2} + 5 = 6 p

x^{2} + 10 x + 16 = 0

r, r^{2} = \frac{1}{θ}

x^{2} - 6.5 x - 1.5

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