What are the solutions to the equation 3y-10+y^2=0 ?

The solutions to the equation 3y-10+y^2=0 are y=2,y=-5

Find the zeros of 3y-10+y^2=0

The zeros of 3y-10+y^2=0 are y=2,y=-5

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3 y - 10 + y^{2} = 0

y = 2, y = - 5

Solution steps

3 y - 10 + y^{2} = 0

Write in the standard form

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

y^{2} + 3 y - 10 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 10) = 7

y_{1, 2} = \frac{- 3 \pm 7}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 3 + 7}{2 \cdot 1}, y_{2} = \frac{- 3 - 7}{2 \cdot 1}

y = \frac{- 3 + 7}{2 \cdot 1} : 2

y = \frac{- 3 - 7}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

y = 2, y = - 5

- 7 - y^{2} = - 25

5 = 200 + 0 \cdot x - 2 (9.8) \cdot x^{2}

x^{2} = \frac{12}{9}

x, (λ^{2} - 25) x^{2} + (λ^{2} - 36) y^{2} = λ

x, x^{2} y - 4 y + x = 0

What are the solutions to the equation 3y-10+y^2=0 ?
The solutions to the equation 3y-10+y^2=0 are y=2,y=-5
Find the zeros of 3y-10+y^2=0
The zeros of 3y-10+y^2=0 are y=2,y=-5