What are the solutions to the equation 25y^2+40y=-16 ?

The solutions to the equation 25y^2+40y=-16 are y=-4/5

Find the zeros of 25y^2+40y=-16

The zeros of 25y^2+40y=-16 are y=-4/5

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25 y^{2} + 40 y = - 16

y = - \frac{4}{5}

Decimal Notation

y = - 0.8

Solution steps

25 y^{2} + 40 y = - 16

Move

16

to the left side

25 y^{2} + 40 y + 16 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 25 \cdot 16}{2 \cdot 25}

4 0^{2} - 4 \cdot 25 \cdot 16 = 0

y_{1, 2} = \frac{- 40 \pm 0}{2 \cdot 25}

y = \frac{- 40}{2 \cdot 25}

\frac{- 40}{2 \cdot 25} = - \frac{4}{5}

y = - \frac{4}{5}

The solution to the quadratic equation is:

y = - \frac{4}{5}

x^{2} - 6 x + 12 = 41 + 12

\frac{2}{5} x^{2} - x = \frac{1}{5}

r, A = π r^{2} h

x, 3 x^{2} + 5 x - 1 = 0

x^{2} - 3 x + 2 = 10

What are the solutions to the equation 25y^2+40y=-16 ?
The solutions to the equation 25y^2+40y=-16 are y=-4/5
Find the zeros of 25y^2+40y=-16
The zeros of 25y^2+40y=-16 are y=-4/5