What are the solutions to the equation A(x)=-x^2+100x ?

The solutions to the equation A(x)=-x^2+100x are x=0,x=100-A

Find the zeros of A(x)=-x^2+100x

The zeros of A(x)=-x^2+100x are x=0,x=100-A

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A (x) = - x^{2} + 100 x

x = 0, x = 100 - A

Solution steps

A (x) = - x^{2} + 100 x

Refine

A x = - x^{2} + 100 x

Switch sides

- x^{2} + 100 x = A x

Move

A x

to the left side

- x^{2} + 100 x - A x = 0

Write in the standard form

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

- x^{2} + (100 - A) x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 100 - A ) \pm ( 100 - A ) ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

Simplify

(100 - A)^{2} - 4 (- 1) \cdot 0 : - A + 100

x_{1, 2} = \frac{- ( 100 - A ) \pm ( - A + 100 )}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( 100 - A ) - A + 100}{2 ( - 1 )}, x_{2} = \frac{- ( 100 - A ) - ( - A + 100 )}{2 ( - 1 )}

x = \frac{- ( 100 - A ) - A + 100}{2 ( - 1 )} : 0

x = \frac{- ( 100 - A ) - ( - A + 100 )}{2 ( - 1 )} : 100 - A

The solutions to the quadratic equation are:

x = 0, x = 100 - A

x^{2} - 24 x + 144 = 144

14 x^{2} + 17 x - 6 = 0

x^{2} - 16 x + 4 = 0

x, x^{2} + 4 x k + 1 = 0

x, y^{2} (x^{2} - 1) - 2 x = 0

What are the solutions to the equation A(x)=-x^2+100x ?
The solutions to the equation A(x)=-x^2+100x are x=0,x=100-A
Find the zeros of A(x)=-x^2+100x
The zeros of A(x)=-x^2+100x are x=0,x=100-A