What are the solutions to the equation y^2-12y+36=0 ?

The solutions to the equation y^2-12y+36=0 are y=6

Find the zeros of y^2-12y+36=0

The zeros of y^2-12y+36=0 are y=6

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y^{2} - 12 y + 36 = 0

y = 6

Solution steps

y^{2} - 12 y + 36 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 36 = 0

y_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 1}

y = \frac{- ( - 12 )}{2 \cdot 1}

\frac{- ( - 12 )}{2 \cdot 1} = 6

y = 6

The solution to the quadratic equation is:

y = 6

- 52

12 5^{2}

(\frac{5}{2})^{- 1}

8 x + 2 = (2 x + 4) 4 x + 5

\frac{11}{9}

What are the solutions to the equation y^2-12y+36=0 ?
The solutions to the equation y^2-12y+36=0 are y=6
Find the zeros of y^2-12y+36=0
The zeros of y^2-12y+36=0 are y=6