What are the solutions to the equation 6(x^2-2)=3(x^2+1) ?

The solutions to the equation 6(x^2-2)=3(x^2+1) are x=sqrt(5),x=-sqrt(5)

Find the zeros of 6(x^2-2)=3(x^2+1)

The zeros of 6(x^2-2)=3(x^2+1) are x=sqrt(5),x=-sqrt(5)

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6 (x^{2} - 2) = 3 (x^{2} + 1)

x = 5, x = - 5

Decimal Notation

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

Solution steps

6 (x^{2} - 2) = 3 (x^{2} + 1)

Expand

6 (x^{2} - 2) : 6 x^{2} - 12

Expand

3 (x^{2} + 1) : 3 x^{2} + 3

6 x^{2} - 12 = 3 x^{2} + 3

Move

12

to the right side

6 x^{2} = 3 x^{2} + 15

Move

3 x^{2}

to the left side

3 x^{2} = 15

Divide both sides by

3

x^{2} = 5

For

x^{2} = f (a)

the solutions are

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

x = 5, x = - 5

x (x + 2) = 120

x, \frac{x ^{2}}{18} = r

s^{2} + 2 s + 20 = 0

y^{2} + 2 y + 5 y = 5 y^{2} - 4 y

x, 6 x^{2} - 6 x y + 6 x = 0

What are the solutions to the equation 6(x^2-2)=3(x^2+1) ?
The solutions to the equation 6(x^2-2)=3(x^2+1) are x=sqrt(5),x=-sqrt(5)
Find the zeros of 6(x^2-2)=3(x^2+1)
The zeros of 6(x^2-2)=3(x^2+1) are x=sqrt(5),x=-sqrt(5)