What are the solutions to the equation 2x^2+2=40 ?

The solutions to the equation 2x^2+2=40 are x=sqrt(19),x=-sqrt(19)

Find the zeros of 2x^2+2=40

The zeros of 2x^2+2=40 are x=sqrt(19),x=-sqrt(19)

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2 x^{2} + 2 = 40

x = 19, x = - 19

Decimal Notation

x = 4.35889 \dots, x = - 4.35889 \dots

Solution steps

2 x^{2} + 2 = 40

Move

2

to the right side

2 x^{2} = 38

Divide both sides by

2

x^{2} = 19

For

x^{2} = f (a)

the solutions are

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

x = 19, x = - 19

- 5 x \leq 25

(- 2)^{7}

- 19 v = - 19

\frac{- 5 + 3 i}{i}

7 + \frac{7}{2}

What are the solutions to the equation 2x^2+2=40 ?
The solutions to the equation 2x^2+2=40 are x=sqrt(19),x=-sqrt(19)
Find the zeros of 2x^2+2=40
The zeros of 2x^2+2=40 are x=sqrt(19),x=-sqrt(19)