What are the solutions to the equation 2y^2-5y-12=0 ?

The solutions to the equation 2y^2-5y-12=0 are y=4,y=-3/2

Find the zeros of 2y^2-5y-12=0

The zeros of 2y^2-5y-12=0 are y=4,y=-3/2

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2 y^{2} - 5 y - 12 = 0

y = 4, y = - \frac{3}{2}

Decimal Notation

y = 4, y = - 1.5

Solution steps

2 y^{2} - 5 y - 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( - 12 )}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 (- 12) = 11

y_{1, 2} = \frac{- ( - 5 ) \pm 11}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 5 ) + 11}{2 \cdot 2}, y_{2} = \frac{- ( - 5 ) - 11}{2 \cdot 2}

y = \frac{- ( - 5 ) + 11}{2 \cdot 2} : 4

y = \frac{- ( - 5 ) - 11}{2 \cdot 2} : - \frac{3}{2}

The solutions to the quadratic equation are:

y = 4, y = - \frac{3}{2}

2 x^{2} + 3 x = 27

p^{2} + 6 p = - 8

N (t) = 4 t - t^{2}

(5 p + 1)^{2} = 8

w^{2} - 3 w - 4 = 0

What are the solutions to the equation 2y^2-5y-12=0 ?
The solutions to the equation 2y^2-5y-12=0 are y=4,y=-3/2
Find the zeros of 2y^2-5y-12=0
The zeros of 2y^2-5y-12=0 are y=4,y=-3/2