What are the solutions to the equation y^2-13y+36=0 ?

The solutions to the equation y^2-13y+36=0 are y=9,y=4

Find the zeros of y^2-13y+36=0

The zeros of y^2-13y+36=0 are y=9,y=4

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y^{2} - 13 y + 36 = 0

y = 9, y = 4

Solution steps

y^{2} - 13 y + 36 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 13)^{2} - 4 \cdot 1 \cdot 36 = 5

y_{1, 2} = \frac{- ( - 13 ) \pm 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 13 ) + 5}{2 \cdot 1}, y_{2} = \frac{- ( - 13 ) - 5}{2 \cdot 1}

y = \frac{- ( - 13 ) + 5}{2 \cdot 1} : 9

y = \frac{- ( - 13 ) - 5}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

y = 9, y = 4

\frac{6 ^{9}}{6 ^{7}}

y^{2} - 8 y + 1 = 0

\frac{\frac{f ^{2} - q ^{2}}{f q}}{\frac{f - q}{q}}

n^{2} + 34 + 225

50 x^{2} - 18

What are the solutions to the equation y^2-13y+36=0 ?
The solutions to the equation y^2-13y+36=0 are y=9,y=4
Find the zeros of y^2-13y+36=0
The zeros of y^2-13y+36=0 are y=9,y=4