What are the solutions to the equation 2x^2=242 ?

The solutions to the equation 2x^2=242 are x=11,x=-11

Find the zeros of 2x^2=242

The zeros of 2x^2=242 are x=11,x=-11

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2 x^{2} = 242

x = 11, x = - 11

Solution steps

2 x^{2} = 242

Divide both sides by

2

x^{2} = 121

For

x^{2} = f (a)

the solutions are

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

x = 121, x = - 121

121 = 11

- 121 = - 11

x = 11, x = - 11

\frac{8 \cdot 1 0 ^{9}}{2 \cdot 1 0 ^{5}}

\frac{2 ( - \frac{3}{2} )}{1 - ( - \frac{3}{2} ) ^{2}}

6 x^{2} + 18 x - 168

- 8 n - 8 = - 72

6 (2 x + 10) \geq 3 x - 9

What are the solutions to the equation 2x^2=242 ?
The solutions to the equation 2x^2=242 are x=11,x=-11
Find the zeros of 2x^2=242
The zeros of 2x^2=242 are x=11,x=-11