What are the solutions to the equation y^2-12y+29=0 ?

The solutions to the equation y^2-12y+29=0 are y=6+sqrt(7),y=6-sqrt(7)

Find the zeros of y^2-12y+29=0

The zeros of y^2-12y+29=0 are y=6+sqrt(7),y=6-sqrt(7)

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y^{2} - 12 y + 29 = 0

y = 6 + 7, y = 6 - 7

Solution steps

y^{2} - 12 y + 29 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 29}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 29 = 27

y_{1, 2} = \frac{- ( - 12 ) \pm 2 7}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 12 ) + 2 7}{2 \cdot 1}, y_{2} = \frac{- ( - 12 ) - 2 7}{2 \cdot 1}

y = \frac{- ( - 12 ) + 2 7}{2 \cdot 1} : 6 + 7

y = \frac{- ( - 12 ) - 2 7}{2 \cdot 1} : 6 - 7

The solutions to the quadratic equation are:

y = 6 + 7, y = 6 - 7

∣ 3 x + 2 ∣ \leq 4

(67 - x) (x + 32) \geq 0

x^{2} + 15 x + 36

2 x - 5 < - 11

x (x - 13) = 0

What are the solutions to the equation y^2-12y+29=0 ?
The solutions to the equation y^2-12y+29=0 are y=6+sqrt(7),y=6-sqrt(7)
Find the zeros of y^2-12y+29=0
The zeros of y^2-12y+29=0 are y=6+sqrt(7),y=6-sqrt(7)