What are the solutions to the equation 4y^2+12y+5=0 ?

The solutions to the equation 4y^2+12y+5=0 are y=-1/2 ,y=-5/2

Find the zeros of 4y^2+12y+5=0

The zeros of 4y^2+12y+5=0 are y=-1/2 ,y=-5/2

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4 y^{2} + 12 y + 5 = 0

y = - \frac{1}{2}, y = - \frac{5}{2}

Decimal Notation

y = - 0.5, y = - 2.5

Solution steps

4 y^{2} + 12 y + 5 = 0

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 4 \cdot 5 = 8

y_{1, 2} = \frac{- 12 \pm 8}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- 12 + 8}{2 \cdot 4}, y_{2} = \frac{- 12 - 8}{2 \cdot 4}

y = \frac{- 12 + 8}{2 \cdot 4} : - \frac{1}{2}

y = \frac{- 12 - 8}{2 \cdot 4} : - \frac{5}{2}

The solutions to the quadratic equation are:

y = - \frac{1}{2}, y = - \frac{5}{2}

- (5 a^{2} b c)^{2} \cdot (5 ab c)^{- 3}

∣ x - 2 ∣ = 7

x^{2} + 14 x + 50 = 0

2 \leq 3 - 2 (x + 1) < 8

9^{2} + 1 2^{2}

What are the solutions to the equation 4y^2+12y+5=0 ?
The solutions to the equation 4y^2+12y+5=0 are y=-1/2 ,y=-5/2
Find the zeros of 4y^2+12y+5=0
The zeros of 4y^2+12y+5=0 are y=-1/2 ,y=-5/2