What are the solutions to the equation 0=2(x-3)^2-4 ?

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

Find the zeros of 0=2(x-3)^2-4

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

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0 = 2 (x - 3)^{2} - 4

x = 2 + 3, x = - 2 + 3

Decimal Notation

x = 4.41421 \dots, x = 1.58578 \dots

Solution steps

0 = 2 (x - 3)^{2} - 4

Switch sides

2 (x - 3)^{2} - 4 = 0

Move

4

to the right side

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

Divide both sides by

2

(x - 3)^{2} = 2

For

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

the solutions are

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

Solve

x - 3 = 2 : x = 2 + 3

Solve

x - 3 = - 2 : x = - 2 + 3

The solutions to the quadratic equation are:

x = 2 + 3, x = - 2 + 3

\frac{7 + 6 i}{6 + i}

4^{\frac{1}{3}} \cdot 1 6^{\frac{1}{3}}

5 x^{2} - 40 x

2 x^{2} - 9 x + 10 = 0

(x - 3)^{2}

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