What are the solutions to the equation 4y^2+20y+25=0 ?

The solutions to the equation 4y^2+20y+25=0 are y=-5/2

Find the zeros of 4y^2+20y+25=0

The zeros of 4y^2+20y+25=0 are y=-5/2

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4 y^{2} + 20 y + 25 = 0

y = - \frac{5}{2}

Decimal Notation

y = - 2.5

Solution steps

4 y^{2} + 20 y + 25 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 25}{2 \cdot 4}

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

y_{1, 2} = \frac{- 20 \pm 0}{2 \cdot 4}

y = \frac{- 20}{2 \cdot 4}

\frac{- 20}{2 \cdot 4} = - \frac{5}{2}

y = - \frac{5}{2}

The solution to the quadratic equation is:

y = - \frac{5}{2}

e^{- 2 l n (3)}

\frac{2}{3} x - \frac{3}{5} = \frac{7}{5}

2 (x^{2} - 10 x y)^{2} - 1152 y^{4}

6 (3)

2 - 6 x = - 16

What are the solutions to the equation 4y^2+20y+25=0 ?
The solutions to the equation 4y^2+20y+25=0 are y=-5/2
Find the zeros of 4y^2+20y+25=0
The zeros of 4y^2+20y+25=0 are y=-5/2