What are the solutions to the equation y^2-16y+64=81 ?

The solutions to the equation y^2-16y+64=81 are y=17,y=-1

Find the zeros of y^2-16y+64=81

The zeros of y^2-16y+64=81 are y=17,y=-1

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y^{2} - 16 y + 64 = 81

y = 17, y = - 1

Solution steps

y^{2} - 16 y + 64 = 81

Move

81

to the left side

y^{2} - 16 y - 17 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot ( - 17 )}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot (- 17) = 18

y_{1, 2} = \frac{- ( - 16 ) \pm 18}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 16 ) + 18}{2 \cdot 1}, y_{2} = \frac{- ( - 16 ) - 18}{2 \cdot 1}

y = \frac{- ( - 16 ) + 18}{2 \cdot 1} : 17

y = \frac{- ( - 16 ) - 18}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

y = 17, y = - 1

r^{2} + 4 r > - 3

∣ - 2 x + 5 ∣ > 3

2 x > 6

\frac{5}{2 - 3}

x^{2} + 3 x - 18 = 0

What are the solutions to the equation y^2-16y+64=81 ?
The solutions to the equation y^2-16y+64=81 are y=17,y=-1
Find the zeros of y^2-16y+64=81
The zeros of y^2-16y+64=81 are y=17,y=-1