Find the Roots (Zeros) y=(3x^2-12)/(8x^2-18x+4)

$y = \frac{3 x^{2} - 12}{8 x^{2} - 18 x + 4}$

Set $\frac{3 x^{2} - 12}{8 x^{2} - 18 x + 4}$ equal to $0$ .

$\frac{3 x^{2} - 12}{8 x^{2} - 18 x + 4} = 0$

Solve for $x$ .

Set the numerator equal to zero.

$3 x^{2} - 12 = 0$

Solve the equation for $x$ .

Add $12$ to both sides of the equation.

$3 x^{2} = 12$

Divide each term in $3 x^{2} = 12$ by $3$ and simplify.

Divide each term in $3 x^{2} = 12$ by $3$ .

$\frac{3 x^{2}}{3} = \frac{12}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x^{2}}{3} = \frac{12}{3}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{12}{3}$

Simplify the right side.

Divide $12$ by $3$ .

$x^{2} = 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

Exclude the solutions that do not make $\frac{3 x^{2} - 12}{8 x^{2} - 18 x + 4} = 0$ true.

$x = - 2$