What are the solutions to the equation 31y+7y^2=20 ?

The solutions to the equation 31y+7y^2=20 are y= 4/7 ,y=-5

Find the zeros of 31y+7y^2=20

The zeros of 31y+7y^2=20 are y= 4/7 ,y=-5

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31 y + 7 y^{2} = 20

y = \frac{4}{7}, y = - 5

Decimal

y = 0.57142 \dots, y = - 5

Solution steps

31 y + 7 y^{2} = 20

Move

20

to the left side

31 y + 7 y^{2} - 20 = 0

Write in the standard form

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

7 y^{2} + 31 y - 20 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 31 \pm 3 1 ^{2} - 4 \cdot 7 ( - 20 )}{2 \cdot 7}

3 1^{2} - 4 \cdot 7 (- 20) = 39

y_{1, 2} = \frac{- 31 \pm 39}{2 \cdot 7}

Separate the solutions

y_{1} = \frac{- 31 + 39}{2 \cdot 7}, y_{2} = \frac{- 31 - 39}{2 \cdot 7}

y = \frac{- 31 + 39}{2 \cdot 7} : \frac{4}{7}

y = \frac{- 31 - 39}{2 \cdot 7} : - 5

The solutions to the quadratic equation are:

y = \frac{4}{7}, y = - 5

x^{2} - 10 x + 23

4 x^{2} - 22 x^{2} - 1 = 0

4 x^{2} - 8 x + 3 = 0

1 6^{2} + 8^{2} = x^{2}

(4 y + 4)^{2} = - 16

What are the solutions to the equation 31y+7y^2=20 ?
The solutions to the equation 31y+7y^2=20 are y= 4/7 ,y=-5
Find the zeros of 31y+7y^2=20
The zeros of 31y+7y^2=20 are y= 4/7 ,y=-5