Find the Roots (Zeros) 4x^2-12x+11=2

$4 x^{2} - 12 x + 11 = 2$

Step 1

Subtract $2$ from both sides of the equation.

$4 x^{2} - 12 x + 11 - 2 = 0$

Step 2

Subtract $2$ from $11$ .

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

Step 3

Factor using the perfect square rule.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

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

Rewrite $9$ as $3^{2}$ .

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

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

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

Rewrite the polynomial.

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

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

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

Step 4

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

$2 x - 3 = 0$

Step 5

Solve for $x$ .

Add $3$ to both sides of the equation.

$2 x = 3$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Step 6