What are the solutions to the equation -5=-(x-2)^2+4 ?

The solutions to the equation -5=-(x-2)^2+4 are x=5,x=-1

Find the zeros of -5=-(x-2)^2+4

The zeros of -5=-(x-2)^2+4 are x=5,x=-1

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- 5 = - (x - 2)^{2} + 4

x = 5, x = - 1

Solution steps

- 5 = - (x - 2)^{2} + 4

Switch sides

- (x - 2)^{2} + 4 = - 5

Move

4

to the right side

- (x - 2)^{2} = - 9

Divide both sides by

- 1

(x - 2)^{2} = 9

For

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

the solutions are

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

Solve

x - 2 = 9 : x = 5

Solve

x - 2 = - 9 : x = - 1

The solutions to the quadratic equation are:

x = 5, x = - 1

2 x (2 x - 1) = 30

4.4 \times 1 0^{- 4} = \frac{x ^{2}}{0.01}

n^{2} + 16 n + 63 = 0

1 5^{2} + 2 5^{2} = c^{2}

8 x^{2} - 8 = 64

What are the solutions to the equation -5=-(x-2)^2+4 ?
The solutions to the equation -5=-(x-2)^2+4 are x=5,x=-1
Find the zeros of -5=-(x-2)^2+4
The zeros of -5=-(x-2)^2+4 are x=5,x=-1