What are the solutions to the equation 3(x^2-4)+7=-2 ?

The solutions to the equation 3(x^2-4)+7=-2 are x=1,x=-1

Find the zeros of 3(x^2-4)+7=-2

The zeros of 3(x^2-4)+7=-2 are x=1,x=-1

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3 (x^{2} - 4) + 7 = - 2

x = 1, x = - 1

Solution steps

3 (x^{2} - 4) + 7 = - 2

Move

7

to the right side

3 (x^{2} - 4) = - 9

Divide both sides by

3

x^{2} - 4 = - 3

Move

4

to the right side

x^{2} = 1

For

x^{2} = f (a)

the solutions are

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

x = 1, x = - 1

1 = 1

- 1 = - 1

x = 1, x = - 1

\frac{x - 5}{( x - 2 ) ( x - 9 )} < 0

\frac{( 15 a ^{5} b ^{3} )}{( 5 a ^{2} b )}

(x + 16) (x - 63) \geq 0

\frac{2 - ( 1 - x )}{3} - x = 1 - \frac{2}{3} x

x^{3} - 125 y^{3}

What are the solutions to the equation 3(x^2-4)+7=-2 ?
The solutions to the equation 3(x^2-4)+7=-2 are x=1,x=-1
Find the zeros of 3(x^2-4)+7=-2
The zeros of 3(x^2-4)+7=-2 are x=1,x=-1