What are the solutions to the equation 5y+3y^2=12 ?

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

Find the zeros of 5y+3y^2=12

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

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5 y + 3 y^{2} = 12

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

Decimal Notation

y = 1.33333 \dots, y = - 3

Solution steps

5 y + 3 y^{2} = 12

Move

12

to the left side

5 y + 3 y^{2} - 12 = 0

Write in the standard form

a x^{2} + b x + c = 0

3 y^{2} + 5 y - 12 = 0

Solve with the quadratic formula

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

5^{2} - 4 \cdot 3 (- 12) = 13

y_{1, 2} = \frac{- 5 \pm 13}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 5 + 13}{2 \cdot 3}, y_{2} = \frac{- 5 - 13}{2 \cdot 3}

y = \frac{- 5 + 13}{2 \cdot 3} : \frac{4}{3}

y = \frac{- 5 - 13}{2 \cdot 3} : - 3

The solutions to the quadratic equation are:

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

y - 6 \geq 12

- 4 x + 28 \geq 3 x + 7

b^{\frac{2}{3}} - b^{\frac{1}{3}}

2 - 2 x = - 2 (x - 1)

\frac{a ^{3}}{a}

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