What are the solutions to the equation 4(x^2-3)+6=10 ?

The solutions to the equation 4(x^2-3)+6=10 are x=2,x=-2

Find the zeros of 4(x^2-3)+6=10

The zeros of 4(x^2-3)+6=10 are x=2,x=-2

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4 (x^{2} - 3) + 6 = 10

x = 2, x = - 2

Solution steps

4 (x^{2} - 3) + 6 = 10

Move

6

to the right side

4 (x^{2} - 3) = 4

Divide both sides by

4

x^{2} - 3 = 1

Move

3

to the right side

x^{2} = 4

For

x^{2} = f (a)

the solutions are

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

x = 4, x = - 4

4 = 2

- 4 = - 2

x = 2, x = - 2

x^{2} - 9 x = 0

y, y^{2} = \frac{x ^{3}}{4 - x}

x^{2} - 9 x + 20 = 0

4 x^{2} + 4 x + 36 = 0

- 5 a^{2} + 5 a + 10 = 0

What are the solutions to the equation 4(x^2-3)+6=10 ?
The solutions to the equation 4(x^2-3)+6=10 are x=2,x=-2
Find the zeros of 4(x^2-3)+6=10
The zeros of 4(x^2-3)+6=10 are x=2,x=-2