What are the solutions to the equation 45y+7y^2=28 ?

The solutions to the equation 45y+7y^2=28 are y= 4/7 ,y=-7

Find the zeros of 45y+7y^2=28

The zeros of 45y+7y^2=28 are y= 4/7 ,y=-7

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45 y + 7 y^{2} = 28

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

Decimal Notation

y = 0.57142 \dots, y = - 7

Solution steps

45 y + 7 y^{2} = 28

Move

28

to the left side

45 y + 7 y^{2} - 28 = 0

Write in the standard form

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

7 y^{2} + 45 y - 28 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 45 \pm 4 5 ^{2} - 4 \cdot 7 ( - 28 )}{2 \cdot 7}

4 5^{2} - 4 \cdot 7 (- 28) = 53

y_{1, 2} = \frac{- 45 \pm 53}{2 \cdot 7}

Separate the solutions

y_{1} = \frac{- 45 + 53}{2 \cdot 7}, y_{2} = \frac{- 45 - 53}{2 \cdot 7}

y = \frac{- 45 + 53}{2 \cdot 7} : \frac{4}{7}

y = \frac{- 45 - 53}{2 \cdot 7} : - 7

The solutions to the quadratic equation are:

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

5 w^{2} - 16 w = 16

m^{2} + 16 m + 73 = 0

v^{2} + 7 v + 12 = 0

4 t^{2} + 19 t - 30 = 0

9 x^{2} - 12 x - 12 = 0

What are the solutions to the equation 45y+7y^2=28 ?
The solutions to the equation 45y+7y^2=28 are y= 4/7 ,y=-7
Find the zeros of 45y+7y^2=28
The zeros of 45y+7y^2=28 are y= 4/7 ,y=-7