What are the solutions to the equation y^2-12y+32=0 ?

The solutions to the equation y^2-12y+32=0 are y=8,y=4

Find the zeros of y^2-12y+32=0

The zeros of y^2-12y+32=0 are y=8,y=4

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y^{2} - 12 y + 32 = 0

y = 8, y = 4

Solution steps

y^{2} - 12 y + 32 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 32 = 4

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

Separate the solutions

y_{1} = \frac{- ( - 12 ) + 4}{2 \cdot 1}, y_{2} = \frac{- ( - 12 ) - 4}{2 \cdot 1}

y = \frac{- ( - 12 ) + 4}{2 \cdot 1} : 8

y = \frac{- ( - 12 ) - 4}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

y = 8, y = 4

(z + 2)^{2} - 5 (z + 2)

12 c + 5 (5 + 3 c) = 3 (9 c - 4)

8 c^{2} - 2 c - 3

10 x^{2} - 20 x = 0

3 3^{2}

What are the solutions to the equation y^2-12y+32=0 ?
The solutions to the equation y^2-12y+32=0 are y=8,y=4
Find the zeros of y^2-12y+32=0
The zeros of y^2-12y+32=0 are y=8,y=4