Find the Roots (Zeros) h(x)=(-2x+3)(-x+3)

$h (x) = (- 2 x + 3) (- x + 3)$

Step 1

Set $(- 2 x + 3) (- x + 3)$ equal to $0$ .

$(- 2 x + 3) (- x + 3) = 0$

Step 2

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$- 2 x + 3 = 0$

$- x + 3 = 0$

Set $- 2 x + 3$ equal to $0$ and solve for $x$ .

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

$- 2 x + 3 = 0$

Solve $- 2 x + 3 = 0$ for $x$ .

Subtract $3$ from 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 $1$ .

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

Simplify the right side.

Dividing two negative values results in a positive value.

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

Set $- x + 3$ equal to $0$ and solve for $x$ .

Set $- x + 3$ equal to $0$ .

$- x + 3 = 0$

Solve $- x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$- x = - 3$

Divide each term in $- x = - 3$ by $- 1$ and simplify.

Divide each term in $- x = - 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 3}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$

The final solution is all the values that make $(- 2 x + 3) (- x + 3) = 0$ true.

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

Step 3

Find the Roots (Zeros) h(x)=(-2x+3)(-x+3)

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