What are the solutions to the equation 2x^2+4x^2=42 ?

The solutions to the equation 2x^2+4x^2=42 are x=sqrt(7),x=-sqrt(7)

Find the zeros of 2x^2+4x^2=42

The zeros of 2x^2+4x^2=42 are x=sqrt(7),x=-sqrt(7)

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2 x^{2} + 4 x^{2} = 42

x = 7, x = - 7

Decimal Notation

x = 2.64575 \dots, x = - 2.64575 \dots

Solution steps

2 x^{2} + 4 x^{2} = 42

Add similar elements:

2 x^{2} + 4 x^{2} = 6 x^{2}

6 x^{2} = 42

Divide both sides by

6

x^{2} = 7

For

x^{2} = f (a)

the solutions are

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

x = 7, x = - 7

7 (2 - x) \geq 28

3 x - 5 > 16

∣ 2 x + 13 ∣ < 15

(\frac{8 x ^{2}}{2 x ^{2}})^{2}

- \frac{1}{4} (2)

What are the solutions to the equation 2x^2+4x^2=42 ?
The solutions to the equation 2x^2+4x^2=42 are x=sqrt(7),x=-sqrt(7)
Find the zeros of 2x^2+4x^2=42
The zeros of 2x^2+4x^2=42 are x=sqrt(7),x=-sqrt(7)