What are the solutions to the equation 0=-(x-3)^2+1 ?

The solutions to the equation 0=-(x-3)^2+1 are x=4,x=2

Find the zeros of 0=-(x-3)^2+1

The zeros of 0=-(x-3)^2+1 are x=4,x=2

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0 = - (x - 3)^{2} + 1

x = 4, x = 2

Solution steps

0 = - (x - 3)^{2} + 1

Switch sides

- (x - 3)^{2} + 1 = 0

Move

1

to the right side

- (x - 3)^{2} = - 1

Divide both sides by

- 1

(x - 3)^{2} = 1

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

x - 3 = 1 : x = 4

Solve

x - 3 = - 1 : x = 2

The solutions to the quadratic equation are:

x = 4, x = 2

10 - 2 (\frac{1}{2})

2 x^{2} - 5 x - 9 = 0

x (x + 4) = 12

20 m^{2} + 13 mn + 2 n^{2}

What are the solutions to the equation 0=-(x-3)^2+1 ?
The solutions to the equation 0=-(x-3)^2+1 are x=4,x=2
Find the zeros of 0=-(x-3)^2+1
The zeros of 0=-(x-3)^2+1 are x=4,x=2