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

The solutions to the equation 2y^2+5y+2=0 are y=-1/2 ,y=-2

Find the zeros of 2y^2+5y+2=0

The zeros of 2y^2+5y+2=0 are y=-1/2 ,y=-2

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2 y^{2} + 5 y + 2 = 0

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

Solution steps

2 y^{2} + 5 y + 2 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

6 x^{2} + 8 x + 2

\frac{3 x - 2}{x + 1} + 4 > 0

7 (5 t - 4) - (2 + 15 t) < 60

\frac{ln ( 2 )}{2.5}

r^{2} - 16 r + 60

What are the solutions to the equation 2y^2+5y+2=0 ?
The solutions to the equation 2y^2+5y+2=0 are y=-1/2 ,y=-2
Find the zeros of 2y^2+5y+2=0
The zeros of 2y^2+5y+2=0 are y=-1/2 ,y=-2