What are the solutions to the equation 5y+3y^2=28 ?

The solutions to the equation 5y+3y^2=28 are y= 7/3 ,y=-4

Find the zeros of 5y+3y^2=28

The zeros of 5y+3y^2=28 are y= 7/3 ,y=-4

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5 y + 3 y^{2} = 28

y = \frac{7}{3}, y = - 4

Decimal Notation

y = 2.33333 \dots, y = - 4

Solution steps

5 y + 3 y^{2} = 28

Move

28

to the left side

5 y + 3 y^{2} - 28 = 0

Write in the standard form

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

3 y^{2} + 5 y - 28 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 3 ( - 28 )}{2 \cdot 3}

5^{2} - 4 \cdot 3 (- 28) = 19

y_{1, 2} = \frac{- 5 \pm 19}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 5 + 19}{2 \cdot 3}, y_{2} = \frac{- 5 - 19}{2 \cdot 3}

y = \frac{- 5 + 19}{2 \cdot 3} : \frac{7}{3}

y = \frac{- 5 - 19}{2 \cdot 3} : - 4

The solutions to the quadratic equation are:

y = \frac{7}{3}, y = - 4

2 a^{2} - 13 a + 6 = 0

10 x^{2} - 9 x - 9 = 0

(x + 5)^{2} = 7 (x + 5)

t, h = t^{2} + 2 t

2 x^{2} - 9 = 12 x

What are the solutions to the equation 5y+3y^2=28 ?
The solutions to the equation 5y+3y^2=28 are y= 7/3 ,y=-4
Find the zeros of 5y+3y^2=28
The zeros of 5y+3y^2=28 are y= 7/3 ,y=-4