What are the solutions to the equation 4y^2+8y+5=0 ?

The solutions to the equation 4y^2+8y+5=0 are y=-1+i 1/2 ,y=-1-i 1/2

Find the zeros of 4y^2+8y+5=0

The zeros of 4y^2+8y+5=0 are y=-1+i 1/2 ,y=-1-i 1/2

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4 y^{2} + 8 y + 5 = 0

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

Solution steps

4 y^{2} + 8 y + 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

Simplify

8^{2} - 4 \cdot 4 \cdot 5 : 4 i

y_{1, 2} = \frac{- 8 \pm 4 i}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- 8 + 4 i}{2 \cdot 4}, y_{2} = \frac{- 8 - 4 i}{2 \cdot 4}

y = \frac{- 8 + 4 i}{2 \cdot 4} : - 1 + i \frac{1}{2}

y = \frac{- 8 - 4 i}{2 \cdot 4} : - 1 - i \frac{1}{2}

The solutions to the quadratic equation are:

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

(12 - 1.9 t)^{2} = 0

t^{2} - 25 t + 144 = 0

(w + 6)^{2} - 40 = 0

x^{2} - 16 x + 13 = 0

(6 x - 1)^{2} = 43

What are the solutions to the equation 4y^2+8y+5=0 ?
The solutions to the equation 4y^2+8y+5=0 are y=-1+i 1/2 ,y=-1-i 1/2
Find the zeros of 4y^2+8y+5=0
The zeros of 4y^2+8y+5=0 are y=-1+i 1/2 ,y=-1-i 1/2