What are the solutions to the equation y^2-8y+16=64 ?

The solutions to the equation y^2-8y+16=64 are y=12,y=-4

Find the zeros of y^2-8y+16=64

The zeros of y^2-8y+16=64 are y=12,y=-4

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y^{2} - 8 y + 16 = 64

y = 12, y = - 4

Solution steps

y^{2} - 8 y + 16 = 64

Move

64

to the left side

y^{2} - 8 y - 48 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot (- 48) = 16

y_{1, 2} = \frac{- ( - 8 ) \pm 16}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 8 ) + 16}{2 \cdot 1}, y_{2} = \frac{- ( - 8 ) - 16}{2 \cdot 1}

y = \frac{- ( - 8 ) + 16}{2 \cdot 1} : 12

y = \frac{- ( - 8 ) - 16}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

y = 12, y = - 4

∣ 3 x ∣ > - 3

∣ 2 x + 8 ∣ = 6 x

- x^{2}

55 (x + 1) = 48 x

55 - 11 u = 55

What are the solutions to the equation y^2-8y+16=64 ?
The solutions to the equation y^2-8y+16=64 are y=12,y=-4
Find the zeros of y^2-8y+16=64
The zeros of y^2-8y+16=64 are y=12,y=-4