What are the solutions to the equation 2(x+4)^2=242 ?

The solutions to the equation 2(x+4)^2=242 are x=7,x=-15

Find the zeros of 2(x+4)^2=242

The zeros of 2(x+4)^2=242 are x=7,x=-15

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2 (x + 4)^{2} = 242

x = 7, x = - 15

Solution steps

2 (x + 4)^{2} = 242

Divide both sides by

2

(x + 4)^{2} = 121

For

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

the solutions are

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

Solve

x + 4 = 121 : x = 7

Solve

x + 4 = - 121 : x = - 15

The solutions to the quadratic equation are:

x = 7, x = - 15

18.82 = \frac{3 x ^{2}}{4}

2 (x - 1)^{2} = 90

(a + 4) x^{2} - 1 = (2 a + 2) x - a

n^{2} - n = - 41

y, x^{3} y^{2} - 5 x^{2} y + x = 1

What are the solutions to the equation 2(x+4)^2=242 ?
The solutions to the equation 2(x+4)^2=242 are x=7,x=-15
Find the zeros of 2(x+4)^2=242
The zeros of 2(x+4)^2=242 are x=7,x=-15