Find the Roots (Zeros) 12x=9x^2+4

$12 x = 9 x^{2} + 4$

Subtract $9 x^{2}$ from both sides of the equation.

$12 x - 9 x^{2} = 4$

Subtract $4$ from both sides of the equation.

$12 x - 9 x^{2} - 4 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $12 x - 9 x^{2} - 4$ .

Reorder $12 x$ and $- 9 x^{2}$ .

$- 9 x^{2} + 12 x - 4 = 0$

Factor $- 1$ out of $- 9 x^{2}$ .

$- (9 x^{2}) + 12 x - 4 = 0$

Factor $- 1$ out of $12 x$ .

$- (9 x^{2}) - (- 12 x) - 4 = 0$

Rewrite $- 4$ as $- 1 (4)$ .

$- (9 x^{2}) - (- 12 x) - 1 \cdot 4 = 0$

Factor $- 1$ out of $- (9 x^{2}) - (- 12 x)$ .

$- (9 x^{2} - 12 x) - 1 \cdot 4 = 0$

Factor $- 1$ out of $- (9 x^{2} - 12 x) - 1 (4)$ .

$- (9 x^{2} - 12 x + 4) = 0$

Factor using the perfect square rule.

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$- ({(3 x)}^{2} - 12 x + 4) = 0$

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

$- ({(3 x)}^{2} - 12 x + 2^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 x = 2 \cdot (3 x) \cdot 2$

Rewrite the polynomial.

$- ({(3 x)}^{2} - 2 \cdot (3 x) \cdot 2 + 2^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 3 x$ and $b = 2$ .

$- {(3 x - 2)}^{2} = 0$

Divide each term in $- {(3 x - 2)}^{2} = 0$ by $- 1$ and simplify.

Divide each term in $- {(3 x - 2)}^{2} = 0$ by $- 1$ .

$\frac{- {(3 x - 2)}^{2}}{- 1} = \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{{(3 x - 2)}^{2}}{1} = \frac{0}{- 1}$

Divide ${(3 x - 2)}^{2}$ by $1$ .

${(3 x - 2)}^{2} = \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

${(3 x - 2)}^{2} = 0$

Set the $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Solve for $x$ .

Add $2$ to both sides of the equation.

$3 x = 2$

Divide each term in $3 x = 2$ by $3$ and simplify.

Divide each term in $3 x = 2$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Find the Roots (Zeros) 12x=9x^2+4

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