What are the solutions to the equation 12y^2+23y+5=0 ?

The solutions to the equation 12y^2+23y+5=0 are y=-1/4 ,y=-5/3

Find the zeros of 12y^2+23y+5=0

The zeros of 12y^2+23y+5=0 are y=-1/4 ,y=-5/3

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12 y^{2} + 23 y + 5 = 0

y = - \frac{1}{4}, y = - \frac{5}{3}

Decimal Notation

y = - 0.25, y = - 1.66666 \dots

Solution steps

12 y^{2} + 23 y + 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 23 \pm 2 3 ^{2} - 4 \cdot 12 \cdot 5}{2 \cdot 12}

2 3^{2} - 4 \cdot 12 \cdot 5 = 17

y_{1, 2} = \frac{- 23 \pm 17}{2 \cdot 12}

Separate the solutions

y_{1} = \frac{- 23 + 17}{2 \cdot 12}, y_{2} = \frac{- 23 - 17}{2 \cdot 12}

y = \frac{- 23 + 17}{2 \cdot 12} : - \frac{1}{4}

y = \frac{- 23 - 17}{2 \cdot 12} : - \frac{5}{3}

The solutions to the quadratic equation are:

y = - \frac{1}{4}, y = - \frac{5}{3}

- \frac{3}{2} x^{- \frac{3}{2}}

12 m^{3} - 14 m^{2} - 30 m + 35

9 (t + 2) = 3 (3 t - 2)

\frac{1}{5} t^{2} - 36 = 0

lo g_{4} (192) - lo g_{4} (3)

What are the solutions to the equation 12y^2+23y+5=0 ?
The solutions to the equation 12y^2+23y+5=0 are y=-1/4 ,y=-5/3
Find the zeros of 12y^2+23y+5=0
The zeros of 12y^2+23y+5=0 are y=-1/4 ,y=-5/3