What are the solutions to the equation solvefor a,y=-a^2+18a ?

The solutions to the equation solvefor a,y=-a^2+18a are a=9-sqrt(-y+81),a=sqrt(-y+81)+9

Find the zeros of solvefor a,y=-a^2+18a

The zeros of solvefor a,y=-a^2+18a are a=9-sqrt(-y+81),a=sqrt(-y+81)+9

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solve for

a, y = - a^{2} + 18 a

a = 9 - - y + 81, a = - y + 81 + 9

Solution steps

y = - a^{2} + 18 a

Switch sides

- a^{2} + 18 a = y

Move

y

to the left side

- a^{2} + 18 a - y = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 1 ) ( - y )}{2 ( - 1 )}

Simplify

1 8^{2} - 4 (- 1) (- y) : 2 81 - y

a_{1, 2} = \frac{- 18 \pm 2 81 - y}{2 ( - 1 )}

Separate the solutions

a_{1} = \frac{- 18 + 2 81 - y}{2 ( - 1 )}, a_{2} = \frac{- 18 - 2 81 - y}{2 ( - 1 )}

a = \frac{- 18 + 2 81 - y}{2 ( - 1 )} : 9 - - y + 81

a = \frac{- 18 - 2 81 - y}{2 ( - 1 )} : - y + 81 + 9

The solutions to the quadratic equation are:

a = 9 - - y + 81, a = - y + 81 + 9

2 x^{2} + 5 x a = - 2 a^{2}

x^{2} - 23 x + 2 = 0

6 x^{2} + 11 x = 7

6 x^{2} + 11 x = 0

x, (x - 3)^{2} = 4 k^{2}

What are the solutions to the equation solvefor a,y=-a^2+18a ?
The solutions to the equation solvefor a,y=-a^2+18a are a=9-sqrt(-y+81),a=sqrt(-y+81)+9
Find the zeros of solvefor a,y=-a^2+18a
The zeros of solvefor a,y=-a^2+18a are a=9-sqrt(-y+81),a=sqrt(-y+81)+9