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

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

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

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

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

x = 4, x = 2

Solution steps

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

Switch sides

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

Move

2

to the right side

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

Divide both sides by

- 2

(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

- 6 = - 36 + 6 n

\frac{1}{3 ^{2}}

j^{2} - 26 j + 7 = 0

\frac{u ^{2 x + 1}}{u ^{x}}

∣ 3 - 7 x ∣ = - 6

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