What are the solutions to the equation 2(x-3)^2+6=14 ?

The solutions to the equation 2(x-3)^2+6=14 are x=5,x=1

Find the zeros of 2(x-3)^2+6=14

The zeros of 2(x-3)^2+6=14 are x=5,x=1

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2 (x - 3)^{2} + 6 = 14

x = 5, x = 1

Solution steps

2 (x - 3)^{2} + 6 = 14

Move

6

to the right side

2 (x - 3)^{2} = 8

Divide both sides by

2

(x - 3)^{2} = 4

For

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

the solutions are

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

Solve

x - 3 = 4 : x = 5

Solve

x - 3 = - 4 : x = 1

The solutions to the quadratic equation are:

x = 5, x = 1

\frac{\frac{4}{3}}{4}

2 x = 5 - 4 x^{2}

x - \frac{5}{2} < \frac{x}{4} - 6

2 - \frac{b}{3} = - \frac{5}{2}

\frac{( \frac{1}{y} + \frac{1}{x} )}{\frac{x}{y} - \frac{y}{x}}

What are the solutions to the equation 2(x-3)^2+6=14 ?
The solutions to the equation 2(x-3)^2+6=14 are x=5,x=1
Find the zeros of 2(x-3)^2+6=14
The zeros of 2(x-3)^2+6=14 are x=5,x=1