What are the solutions to the equation 2y^2+10y=28 ?

The solutions to the equation 2y^2+10y=28 are y=2,y=-7

Find the zeros of 2y^2+10y=28

The zeros of 2y^2+10y=28 are y=2,y=-7

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2 y^{2} + 10 y = 28

y = 2, y = - 7

Solution steps

2 y^{2} + 10 y = 28

Move

28

to the left side

2 y^{2} + 10 y - 28 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 2 ( - 28 )}{2 \cdot 2}

1 0^{2} - 4 \cdot 2 (- 28) = 18

y_{1, 2} = \frac{- 10 \pm 18}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 10 + 18}{2 \cdot 2}, y_{2} = \frac{- 10 - 18}{2 \cdot 2}

y = \frac{- 10 + 18}{2 \cdot 2} : 2

y = \frac{- 10 - 18}{2 \cdot 2} : - 7

The solutions to the quadratic equation are:

y = 2, y = - 7

\frac{- \frac{3}{4} + \frac{8}{15}}{1 - ( - \frac{3}{4} ) ( \frac{8}{15} )}

3 x^{3} > 36 x - 3 x^{2}

y^{2} - 4 y + 5 = 0

\frac{z}{12} = \frac{4}{5}

(4 x^{2} y^{4}) (2 x^{5} y)

What are the solutions to the equation 2y^2+10y=28 ?
The solutions to the equation 2y^2+10y=28 are y=2,y=-7
Find the zeros of 2y^2+10y=28
The zeros of 2y^2+10y=28 are y=2,y=-7