What are the solutions to the equation 10+2x^2=20 ?

The solutions to the equation 10+2x^2=20 are x=sqrt(5),x=-sqrt(5)

Find the zeros of 10+2x^2=20

The zeros of 10+2x^2=20 are x=sqrt(5),x=-sqrt(5)

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10 + 2 x^{2} = 20

x = 5, x = - 5

Decimal

x = 2.23606 \dots, x = - 2.23606 \dots

Solution steps

10 + 2 x^{2} = 20

Move

10

to the right side

2 x^{2} = 10

Divide both sides by

2

x^{2} = 5

For

x^{2} = f (a)

the solutions are

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

x = 5, x = - 5

- 36 + 20 x^{4}

15 \cdot 8

∣ - 8 n ∣ < 32

x^{2} = 5 x - 6

(x)^{\frac{3}{2}}

What are the solutions to the equation 10+2x^2=20 ?
The solutions to the equation 10+2x^2=20 are x=sqrt(5),x=-sqrt(5)
Find the zeros of 10+2x^2=20
The zeros of 10+2x^2=20 are x=sqrt(5),x=-sqrt(5)