Find the Roots (Zeros) 9y^2-30y+25=0

$9 y^{2} - 30 y + 25 = 0$

Step 1

Solve for $y$ .

Factor using the perfect square rule.

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

${(3 y)}^{2} - 30 y + 25 = 0$

Rewrite $25$ as $5^{2}$ .

${(3 y)}^{2} - 30 y + 5^{2} = 0$

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

$30 y = 2 \cdot (3 y) \cdot 5$

Rewrite the polynomial.

${(3 y)}^{2} - 2 \cdot (3 y) \cdot 5 + 5^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 3 y$ and $b = 5$ .

${(3 y - 5)}^{2} = 0$

Set the $3 y - 5$ equal to $0$ .

$3 y - 5 = 0$

Solve for $y$ .

Add $5$ to both sides of the equation.

$3 y = 5$

Divide each term in $3 y = 5$ by $3$ and simplify.

Divide each term in $3 y = 5$ by $3$ .

$\frac{3 y}{3} = \frac{5}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{5}{3}$

Divide $y$ by $1$ .

$y = \frac{5}{3}$

Step 2

Set $0$ equal to $0$ .

$\frac{5}{3} = 0$

Step 3

Solve for $x$ .

Set the numerator equal to zero.

$5 = 0$

Since $5 \neq 0$ , there are no solutions.

No solution