What are the solutions to the equation 20y^2+39y+18=0 ?

The solutions to the equation 20y^2+39y+18=0 are y=-3/4 ,y=-6/5

Find the zeros of 20y^2+39y+18=0

The zeros of 20y^2+39y+18=0 are y=-3/4 ,y=-6/5

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20 y^{2} + 39 y + 18 = 0

y = - \frac{3}{4}, y = - \frac{6}{5}

Decimal Notation

y = - 0.75, y = - 1.2

Solution steps

20 y^{2} + 39 y + 18 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 39 \pm 3 9 ^{2} - 4 \cdot 20 \cdot 18}{2 \cdot 20}

3 9^{2} - 4 \cdot 20 \cdot 18 = 9

y_{1, 2} = \frac{- 39 \pm 9}{2 \cdot 20}

Separate the solutions

y_{1} = \frac{- 39 + 9}{2 \cdot 20}, y_{2} = \frac{- 39 - 9}{2 \cdot 20}

y = \frac{- 39 + 9}{2 \cdot 20} : - \frac{3}{4}

y = \frac{- 39 - 9}{2 \cdot 20} : - \frac{6}{5}

The solutions to the quadratic equation are:

y = - \frac{3}{4}, y = - \frac{6}{5}

8^{- \frac{4}{3}}

2 (3 x + 1)^{2} = 8

\frac{x ^{5} y ^{2}}{x y ^{2}}

ln (7 x)

2 < \frac{3 - 3 x}{12} < 6

What are the solutions to the equation 20y^2+39y+18=0 ?
The solutions to the equation 20y^2+39y+18=0 are y=-3/4 ,y=-6/5
Find the zeros of 20y^2+39y+18=0
The zeros of 20y^2+39y+18=0 are y=-3/4 ,y=-6/5