What are the solutions to the equation -y^2+10y=24 ?

The solutions to the equation -y^2+10y=24 are y=4,y=6

Find the zeros of -y^2+10y=24

The zeros of -y^2+10y=24 are y=4,y=6

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- y^{2} + 10 y = 24

y = 4, y = 6

Solution steps

- y^{2} + 10 y = 24

Move

24

to the left side

- y^{2} + 10 y - 24 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 1 ) ( - 24 )}{2 ( - 1 )}

1 0^{2} - 4 (- 1) (- 24) = 2

y_{1, 2} = \frac{- 10 \pm 2}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- 10 + 2}{2 ( - 1 )}, y_{2} = \frac{- 10 - 2}{2 ( - 1 )}

y = \frac{- 10 + 2}{2 ( - 1 )} : 4

y = \frac{- 10 - 2}{2 ( - 1 )} : 6

The solutions to the quadratic equation are:

y = 4, y = 6

x^{2} + 6 x + 3 = 80

4 x^{2} + 28 x + 49 = 5

10 x^{2} + 10 x = 0

34.04 = 0.9 \cdot 2 \cdot 0.85 \cdot 210 \cdot x \cdot 0.2 (40 - \frac{x}{2})

34 x + x^{2}

What are the solutions to the equation -y^2+10y=24 ?
The solutions to the equation -y^2+10y=24 are y=4,y=6
Find the zeros of -y^2+10y=24
The zeros of -y^2+10y=24 are y=4,y=6